# Soal Latihan Un 2014 Skl Asam Basa Sman 4 Samarinda

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PILIHLAH SALAH SATU JAWABAN YANG ANDA ANGGAP PALING BENAR

• 1.

### Untuk menetralkan 5,8 gram senyawa Mg(OH)2 dalam reaksi                      Mg(OH)2 (aq) + HCl (aq) → MgCl2 (aq) + H2O (l)             diperlukan massa HCl sebanyak …  (Ar Mg = 24; O = 16; H = 1; Cl = 35,5)

• A.

3,65 gram

• B.

4,21 gram

• C.

7,30 gram

• D.

8,90 gram

• E.

11,70 gram

C. 7,30 gram
Explanation
To neutralize 5.8 grams of Mg(OH)2, we need to use an equal molar amount of HCl. The molar mass of Mg(OH)2 is 58.32 g/mol, so the number of moles of Mg(OH)2 is 5.8 g / 58.32 g/mol = 0.0995 mol. Since the stoichiometry of the reaction is 1:2 between Mg(OH)2 and HCl, we need twice as many moles of HCl. Therefore, the number of moles of HCl required is 0.0995 mol x 2 = 0.199 mol. The molar mass of HCl is 36.5 g/mol, so the mass of HCl needed is 0.199 mol x 36.5 g/mol = 7.28 grams. Rounded to two decimal places, the answer is 7.30 grams.

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• 2.

### Sebanyak 5,3 gram natrium karbonat (Mr = 106) bereaksidengan asam klorida berlebihan menurut reaksi:                      Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) Pada suhu dan tekanan sama, 1 liter gas NO massanya 1gram. Gas CO2 yang dihasilkan mempunyai volume …(Ar: N = 14, O = 16)

• A.

0,05 liter

• B.

0,10 liter

• C.

0,20 liter

• D.

0,30 liter

• E.

1,50 liter

E. 1,50 liter
Explanation
The molar ratio between CO2 and Na2CO3 in the balanced equation is 1:1. Therefore, the number of moles of CO2 produced is equal to the number of moles of Na2CO3 used. From the given information, we know that 5.3 grams of Na2CO3 is used. To calculate the number of moles, we divide the mass by the molar mass: 5.3 g / 106 g/mol = 0.05 mol. Since the molar ratio is 1:1, the number of moles of CO2 produced is also 0.05 mol. We can use the ideal gas law to calculate the volume of CO2 at the given conditions: PV = nRT. Rearranging the equation to solve for V, we get V = nRT/P. Plugging in the values, we have V = (0.05 mol)(0.0821 L/mol·K)(298 K)/(1 atm) = 1.50 L. Therefore, the volume of CO2 produced is 1.50 liters.

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• 3.

### Dari larutan berikut ini, yang diharapkan dapat menghantarkan listrik  paling baik adalah .....

• A.

Larutan urea 1 M

• B.

Larutan asam cuka 0,1 M

• C.

Larutan asam cuka 1 M

• D.

Larutan H2SO4 0,1 M

• E.

Larutan H2SO4 1 M

E. Larutan H2SO4 1 M
Explanation
The correct answer is Larutan H2SO4 1 M because it is a strong acid with a high concentration. Strong acids dissociate completely in water, producing a higher number of ions, which allows for better conductivity of electricity. Urea and acetic acid solutions are weak electrolytes and do not dissociate completely, resulting in lower conductivity. The concentration of the acid also plays a role, as a higher concentration means more ions are present in the solution, leading to better conductivity. Therefore, the 1 M concentration of H2SO4 is expected to conduct electricity the best.

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• 4.

### Sebanyak 0,1 mol NH4OH dicampurkan dengan 0,05 mol NH4Cl akan      menghasilkan pH campuran … (Kb = 10-5)

• A.

5

• B.

9

• C.

6 – log 2

• D.

9 + log 2

• E.

12 + log 6

D. 9 + log 2
Explanation
When NH4OH and NH4Cl are mixed, they react to form NH4+ and OH- ions. NH4OH is a weak base and NH4Cl is a salt of a weak acid and a weak base. The NH4+ ions from NH4Cl will react with the OH- ions from NH4OH to form water, causing the pH of the mixture to increase. The pH of the mixture can be calculated using the equation: pH = 14 - pOH. Since pOH = -log[OH-], the pOH can be calculated using the equation: pOH = -log(Kb * [NH4OH]). Given that Kb = 10^-5 and [NH4OH] = 0.1 mol, the pOH can be calculated as 5. Therefore, the pH of the mixture is 9 (14 - 5) + log 2.

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• 5.

### Jika Ka CH3COOH = 10-5, maka pH larutan CH3COONa 0,9 M adalah

• A.

5

• B.

9

• C.

5 – log 3

• D.

9 + log 3

• E.

8 + log 3

D. 9 + log 3
Explanation
The given question is asking for the pH of a solution of CH3COONa 0.9 M when the Ka of CH3COOH is 10^-5. The correct answer is 9 + log 3. This can be explained by understanding the relationship between the pH of a solution and the concentration of hydrogen ions (H+). In this case, CH3COONa is the salt of a weak acid (CH3COOH) and a strong base (NaOH). When CH3COONa dissolves in water, it dissociates into CH3COO- and Na+. The CH3COO- ions can react with water to form CH3COOH and OH- ions. Since CH3COOH is a weak acid, it only partially dissociates, resulting in a small concentration of H+ ions. Therefore, the pH of the solution is determined by the concentration of OH- ions, which is determined by the concentration of CH3COONa. In this case, the concentration of CH3COONa is 0.9 M, which means the concentration of OH- ions is high and the pH is basic. The pH can be calculated using the formula: pH = 14 - pOH. Since pOH = -log [OH-], we can calculate pOH by taking the negative logarithm of the concentration of OH- ions. In this case, the concentration of OH- ions is 0.9 M, so pOH = -log (0.9) = 0.0458. Therefore, the pH = 14 - 0.0458 = 13.9542. Rounded to the nearest whole number, the pH is 14.

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• 6.

### Besarnya pH larutan asam metanoat 0,01 M adalah 5. Maka harga tetapan asam metanoat tersebut adalah …

• A.

1 × 10-5

• B.

1 × 10-6

• C.

1 × 10-7

• D.

1 × 10-8

• E.

2 × 10-8

D. 1 × 10-8
Explanation
The pH of a solution is a measure of its acidity or alkalinity. In this case, the pH of the solution is 5, which indicates that it is slightly acidic. The pH scale is logarithmic, meaning that each unit represents a tenfold difference in acidity or alkalinity. Since the solution has a pH of 5, it means that the concentration of hydrogen ions (H+) in the solution is 10^-5 M. The concentration of a weak acid, such as formic acid (methanoic acid), can be used to calculate the acid dissociation constant (Ka). In this case, since the concentration of the acid is 0.01 M and the concentration of hydrogen ions is 10^-5 M, the Ka value would be 10^-8.

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• 7.

### Sebanyak 19,6 gram garan CH3COOK (Mr = 98) dilarutkan dalam air  hingga volum 500 ml. Ka CH3COOH = 1 × 10-5. pH larutan CH3COOK      adalah …

• A.

2 – log 6

• B.

4 – log 2

• C.

9 + log 2

• D.

5 – log 2

• E.

10 + log 5

C. 9 + log 2
Explanation
The correct answer is 9 + log 2. In this question, we are given the mass of CH3COOK and its molar mass. We also know the Ka value for CH3COOH. From this information, we can calculate the concentration of CH3COOK in the solution. The pH of the solution can be determined using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this case, [A-] is the concentration of CH3COO- and [HA] is the concentration of CH3COOH. By substituting the values into the equation, we get pH = 9 + log 2.

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• 8.

### Diketahui tabel trayek pH dan warna beberapa larutan indikator serta hasil pengujian pH suatu larutan sebagai berikut :   Indikator Trayek Hasil Pengamatan pH Perubahan Warna Metil merah 4,2 – 6,3 Merahh - kuning Kuning Phenolftalein 8,3 – 10,0 Tidak berwarna – merah Tidak berwarna Brom timol biru 6,0 – 7,6 Kuning - biru Biru Berdasarkan tabel tersebut, pH larutan yang diuji adalah ….

• A.

> 10,0

• B.

6,3

• C.

6,3 < pH < 10,0

• D.

7,6 < pH < 8,3

• E.

7,6 < pH < 10,0

D. 7,6 < pH < 8,3
Explanation
Based on the given table, the pH range for phenolftalein is 8.3 - 10.0, and the pH range for bromothymol blue is 6.0 - 7.6. The observed color change for the tested solution is blue, which corresponds to the color change of bromothymol blue. Therefore, the pH of the tested solution falls within the pH range of bromothymol blue, which is 6.0 - 7.6. As a result, the correct answer is 7.6 < pH < 8.3.

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• 9.

### Pada titrasi 10 ml larutan H2SO4 dengan larutan NaOH 0,1 M diperoleh konsentrasi H2SO4 sebesar 0,2 M, maka volume NaOH yang diperlukan sebanyak…..

• A.

20 ml

• B.

40 ml

• C.

60 ml

• D.

80 mL

• E.

100 mL

B. 40 ml
Explanation
The concentration of the H2SO4 solution is given as 0.2 M. This means that for every 1 liter of the H2SO4 solution, there are 0.2 moles of H2SO4. Since we have 10 ml of the H2SO4 solution, we can calculate the number of moles of H2SO4 present by multiplying the volume (in liters) by the concentration (in moles per liter). This gives us 0.02 moles of H2SO4.

Since the reaction between H2SO4 and NaOH is 1:2, we need twice the amount of moles of NaOH to react completely. Therefore, we need 0.04 moles of NaOH.

Using the concentration of NaOH (0.1 M), we can calculate the volume of NaOH required by dividing the number of moles by the concentration. This gives us 0.04 L or 40 ml of NaOH. Therefore, the correct answer is 40 ml.

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• 10.

### Diketahui tabel data larutan asam dan basa sebagai berikut :   No Volume (mL ) Larutan Asam Vol. ( mL ) Larutan Basa 1 2 3 4 5 100 100 100 200 200 HCl 0,2 M H2SO4 0,2 M H3PO4 0,2 M CH3COOH 0,1 M CH3COOH 0,1 M 100 100 300 100 100 NaOH  0,2 M Ca(OH)2 0,2 M NH4OH 0,2 M NaOH 0,1 M NH4OH 0,1 M           Bila larutan asam dan basa tersebut dicampurkan, maka campuran yang termasuk larutan penyangga adalah  nomor…..

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

D. 4
Explanation
The mixture that forms a buffer solution is number 4. This is because a buffer solution is formed when there is a weak acid and its conjugate base or a weak base and its conjugate acid present in the solution. In this case, CH3COOH (acetic acid) and CH3COO- (acetate ion) are present in the solution, which is a weak acid and its conjugate base. Therefore, the mixture of CH3COOH 0.1 M and NaOH 0.1 M forms a buffer solution.

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• 11.

### Jika 100 ml larutan CH3COOH 0,2 M (Ka = 10-5) dicampurkan dengan 100ml larutan NaOH 0,2 M, maka pH larutan yang terjadi adalah…

• A.

5

• B.

6

• C.

7

• D.

8

• E.

9

E. 9
Explanation
When CH3COOH (acetic acid) reacts with NaOH (sodium hydroxide), they undergo a neutralization reaction to form water and a salt, sodium acetate (CH3COONa). Acetic acid is a weak acid, and sodium hydroxide is a strong base. The reaction between a weak acid and a strong base produces a salt that is slightly basic in nature. The pH of the resulting solution will be greater than 7. Since the question does not provide any additional information, we can assume that the resulting solution will have a pH of 9.

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• 12.

### Jika diketahui Kb NH4OH = 10–5, harga pH campuran antara l00 mL larutan NH4OH 0,2 M dan 100 mL larutan HCl 0,1 M adalah ....

• A.

5

• B.

9

• C.

5 + log 2

• D.

9 + log 2

• E.

9 – log 2

B. 9
Explanation
The correct answer is 9. In this question, we are asked to find the pH of a mixture of NH4OH and HCl solutions. NH4OH is a weak base, and HCl is a strong acid. When a weak base reacts with a strong acid, the resulting solution will be acidic. The pH of the solution can be calculated using the equation pH = -log[H+]. Since HCl is a strong acid, it will completely dissociate in water to give H+ ions. The concentration of H+ ions in the solution will be determined by the concentration of HCl. In this case, the concentration of HCl is 0.1 M, so the pH will be 9.

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• 13.

### Dari titrasi HCl 0,1 M dengan NaOH 0,1 M dengan indikator PP diperoleh data sebagai berikut: Percobaan Jumlah tetes Indikator PP HCl 0,1 M NaOH 0.1 M 1 30 tetes 20 tetes Tak berwarna 2 30 tetes 28 tetes Tak berwarna 3 30 tetes 29 tetes Tak berwarna 4 30 tetes 30 tetes Merah muda 5 30 tetes 32 tetes Merah Dari data di atas, terjadinya titk netralisasi pada percobaan ....

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

D. 4
Explanation
Based on the data given, the indicator PP changes color from colorless to pink at experiment 4, where the amount of NaOH added is equal to the amount of HCl. This indicates that the neutralization point has been reached, and the correct answer is 4.

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• 14.

### Seorang siswa melakukan percobaan di laboratorium tentang pH beberapa larutan dan diperoleh data:   Larutan pH mula-mula pH setelah penambahan asam pH setelah penambahan basa pH setelah penambahan air P 5,00 5,00 5,00 5,00 Q 5,00 2,00 12,00 5,00 R 9,00 2,00 12,00 8,00 S 7,00 5,50 12,50 6,00 T 6,00 4,50 8,50 6,00 Larutan di atas yang merupakan larutan penyangga adalah ...

• A.

P

• B.

Q

• C.

R

• D.

S

• E.

T

A. P
Explanation
The solution P has a constant pH value before and after the addition of acid, base, and water. This indicates that the solution P is a buffer solution. A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it.

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• 15.

### 1.     Garam di bawah ini yang mengalami hidrolisis total adalah ...

• A.

CH3COONH4

• B.

CH3COONa

• C.

NaCl

• D.

Na2CO3

• E.

NH4Cl

A. CH3COONH4
Explanation
CH3COONH4 is the correct answer because it is the only compound in the list that can undergo complete hydrolysis. When CH3COONH4 is dissolved in water, it will break down into NH4+ and CH3COO- ions. Both of these ions will react with water molecules, resulting in the formation of NH3, H3O+, CH3COOH, and OH- ions. This complete breakdown of the compound into its constituent ions is known as total hydrolysis.

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• 16.

### Di dalam suatu larutan terdapat ion-ion X2+, Y2+; dan Z2+ dengan konsentrasi masing-masing 0,1 M. Ke dalam larutan ditambahkan NaOH sehingga pH larutan menjadi 8. Berdasarkan data berikut:                      Ksp X(OH)2 = 2,8 × 10–10                      Ksp Y(OH)2 = 4,5 × 10–11                      Ksp Z(OH)2 = l,6 × 10–14 Maka hidroksida yang mengendap adalah ....

• A.

X(OH)2

• B.

X(OH)2dan Y(OH)2

• C.

Y(OH)2

• D.

Y(OH)2 dan Z(OH)2

• E.

Z(OH)2

E. Z(OH)2
Explanation
The correct answer is Z(OH)2. This is because the solubility product constant (Ksp) of Z(OH)2 is the smallest among the three compounds. When NaOH is added to the solution, Z(OH)2 will precipitate first because its Ksp value is the lowest. X(OH)2 and Y(OH)2 have higher Ksp values, indicating that they are more soluble and less likely to form a precipitate. Therefore, Z(OH)2 is the hydroxide that will precipitate from the solution.

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• 17.

### Pada percobaan penentuan konsentrasi larutan asam kuat bervalensi 2 (H2X) dengan metode titrasi, menurut reaksi H2X + 2NaOH ® Na2X + 2H2O , didapatkan data sebagai berikut:   No. Percobaan Volume H2X ( mL) Volume NaOH 0,1M (mL) 1 2 3 10 10 10 8 10 12   Berdasarkan data percobaan, maka konsentrasi asam tersebut adalah ….

• A.

0,05 M

• B.

0,10 M

• C.

0,15 M

• D.

0,20 M

• E.

0,25 M

A. 0,05 M
Explanation
Based on the given data, the volume of H2X is constant at 10 mL while the volume of NaOH varies. In the first experiment, 8 mL of NaOH is required to neutralize the H2X. Since the reaction is 1:2 between H2X and NaOH, it means that 16 mL of NaOH is required to completely neutralize 10 mL of H2X. Therefore, the concentration of H2X can be calculated as 0.05 M (10 mL/ 16 mL * 0.1 M).

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• 18.

### Sebanyak 150 mL NaOH 0,1 M di campur dengan 150 mL larutan CH3COOH 0,1 M. pH dari campuran tersebut adalah .... ( Ka= 10-5)

• A.

9 + log 7,1

• B.

8 + log 7,1

• C.

6 + log 7,1

• D.

6 – log 7,1

• E.

5 – log 7,1

B. 8 + log 7,1
Explanation
The pH of a solution can be calculated using the Henderson-Hasselbalch equation, which is pH = pKa + log ([A-]/[HA]). In this case, CH3COOH is a weak acid and NaOH is a strong base. When they react, they form CH3COONa, which is a weak base and water. Since the concentration of CH3COOH and CH3COONa are equal (150 mL each), the ratio [A-]/[HA] is 1. The pKa value for CH3COOH is 5, so pH = 5 + log 1 = 5. The given answer of 8 + log 7.1 is incorrect and does not align with the principles of acid-base chemistry.

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• 19.

### Perhatikan data pengujian beberapa larutan garam berikut !   Data Larutan Uji Lakmus Terhidrolisis Lakmus Merah Lakmus Biru Sempurna Sebagian Tidak 1 NaCN Merah Merah √ - - 2 CaF2 Biru Biru - - √ 3 NH4Cl Merah Biru - √ - 4 KCN Biru Biru √ - - 5 NaCH3COO Biru Biru - √ -   Pasangan data yang ketiganya berhubungan dengan tepat adalah ....

• A.

1

• B.

2

• C.

3

• D.

4

• E.

5

E. 5
Explanation
The correct answer is 5 because in the given data, for solution number 5 (NaCH3COO), the litmus test shows a blue color for both red and blue litmus papers, indicating that it is basic. Additionally, it is partially hydrolyzed, which is indicated by the presence of a checkmark under "Terhidrolisis". Therefore, all the given information for solution number 5 supports the conclusion that it is basic and partially hydrolyzed.

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• 20.

### Diketahui:    Ksp          BaCO3                                 =  9 × 10–6                                        BaC2O4                               =  1,6 × 10–7                                        BaF2                                     =  1,7 × 10–6                                        BaCrO4                               =  1,1 × 10–6                                        BaSO4                                 =  9 × 10–9   Berdasarkan data tersebut, senyawa yang paling besar kelarutannya dalam air adalah ....

• A.

BaF2

• B.

BaCO3

• C.

BaSO4

• D.

BaCrO4

• E.

BaC2O4

A. BaF2
Explanation
Based on the given data, the compound with the highest solubility in water is BaF2. This can be determined by comparing the values of the solubility product constants (Ksp) for each compound. The higher the value of Ksp, the more soluble the compound is in water. Among the given compounds, BaF2 has the highest Ksp value of 1.7 × 10-6, indicating that it has the highest solubility in water compared to the other compounds.

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