Remed Uts Genap Xi-IPA-2 Smanio (The Number Of Students Absent 1-10)
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Questions and Answers
- 1.
Dalam sebuah percobaan di laboratorium, seorang siswa melakukan titrasi 10 ml larutan HCl dengan larutan NaOH 0,1 M. Ternyata dibutuhkan 20 ml NaOH sampai larutan menjadi berwarna merah maka molaritas HCl adalah …..
- A.
0,002 M
- B.
0,02 M
- C.
0,2 M
- D.
2 M
- E.
20 M
Correct Answer
C. 0,2 MExplanation
The molaritas HCl is 0.2 M because it took 20 ml of 0.1 M NaOH to reach the endpoint of the titration. The balanced chemical equation for the reaction between HCl and NaOH is HCl + NaOH -> NaCl + H2O. From the equation, we can see that the ratio of HCl to NaOH is 1:1. Therefore, if it took 20 ml of 0.1 M NaOH to neutralize 10 ml of HCl, then the molaritas of HCl must be twice the molaritas of NaOH, which is 0.2 M.Rate this question:
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- 2.
Jika ke dalam 50 mL larutan penyangga dengan pH = 5 ditambahkan 50 mLakuades, maka ... .
- A.
PH akan naik sedikit
- B.
PH akan turun sedikit
- C.
PH tidak berbah
- D.
PH naik drastis
- E.
PH turun drastis
Correct Answer
C. PH tidak berbahExplanation
When 50 mL of distilled water is added to a 50 mL buffer solution with a pH of 5, the pH will not change significantly. This is because a buffer solution is able to resist changes in pH when small amounts of acid or base are added. In this case, the buffer solution is able to maintain its pH of 5 even after the addition of distilled water.Rate this question:
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- 3.
Lakmus biru akan menjadi merah apabila dicelupkan dalam larutan … .
- A.
NaOH
- B.
Ba(NO3)2
- C.
(NH4)2SO4
- D.
K2CO3
- E.
CaCl2
Correct Answer
C. (NH4)2SO4Explanation
Lakmus biru akan menjadi merah apabila dicelupkan dalam larutan (NH4)2SO4.Rate this question:
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- 4.
Sebanyak 50 mL larutan CH3COOH 0,1 M (Ka = 10–5) direaksikan dengan 50 mL larutan KOH 0,1 M. pH campuran yang terjadi adalah …
- A.
3
- B.
6-log 7
- C.
8 + log 7
- D.
9-log 7
- E.
9 + log 7
Correct Answer
C. 8 + log 7Explanation
The given question involves the reaction between CH3COOH (acetic acid) and KOH (potassium hydroxide). Acetic acid is a weak acid with a dissociation constant (Ka) of 10^-5. When acetic acid reacts with potassium hydroxide, it forms water and potassium acetate. The pH of the resulting mixture can be calculated using the equation pH = pKa + log ([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. In this case, since the initial concentrations of acetic acid and potassium hydroxide are the same (0.1 M), the resulting mixture will have a pH equal to pKa, which is approximately 8. Adding the logarithm of 7 to 8 gives the answer of 8 + log 7.Rate this question:
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- 5.
Dari garam berikut, yang mengalami hidrolisis total adalah … .
- A.
NH4Br
- B.
K2CO3
- C.
BaCO3
- D.
AlCl3
- E.
Al2(CO3)3
Correct Answer
E. Al2(CO3)3Explanation
Al2(CO3)3 is the correct answer because it is a salt that contains a metal cation (Al3+) and a polyatomic anion (CO32-). When this compound is dissolved in water, it undergoes complete hydrolysis, meaning that it reacts with water to produce hydroxide ions (OH-) and a weak acid. The hydroxide ions make the solution basic, while the weak acid contributes to the acidic nature of the solution. Therefore, Al2(CO3)3 is the only salt among the options that undergoes total hydrolysis.Rate this question:
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- 6.
Untuk menetralkan 20 ml H2SO4 0,2 M dengan cara titrasi diperlukan KOH 0,1 M sebanyak...
- A.
10 ml
- B.
20 ml
- C.
30 ml
- D.
40 ml
- E.
80 ml
Correct Answer
E. 80 mlExplanation
To neutralize 20 ml of 0.2 M H2SO4 using titration, an equal amount of KOH is needed. Since the concentration of KOH is 0.1 M, the volume of KOH required can be calculated using the formula: M1V1 = M2V2. Rearranging the formula, we have V2 = (M1V1) / M2. Plugging in the values, we get V2 = (0.2 M * 20 ml) / 0.1 M = 40 ml. Therefore, 40 ml of 0.1 M KOH is needed to neutralize 20 ml of 0.2 M H2SO4.Rate this question:
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- 7.
12 gram magnesium dimasukkan dalam larutan HCl sehingga habis bereaksi. Mg +2HCl → MgCl2 + H2 Jika gas hidrogen yang terjadi diukur pada suhu dan tekanan standar (STP), maka volume gas hidrogen tersebut adalah..... (Ar Mg = 24)
- A.
22,4 L
- B.
11,2 L
- C.
4,48 L
- D.
2,24 L
- E.
1,12 L
Correct Answer
B. 11,2 LExplanation
When 12 grams of magnesium reacts with hydrochloric acid (HCl), it produces magnesium chloride (MgCl2) and hydrogen gas (H2). According to the balanced chemical equation, 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. Since the molar mass of magnesium is 24 g/mol, we can calculate that 12 grams of magnesium is equal to 0.5 moles. Therefore, the volume of hydrogen gas produced can be calculated using the ideal gas law at STP (standard temperature and pressure). At STP, 1 mole of any gas occupies 22.4 liters. Hence, 0.5 moles of hydrogen gas will occupy 11.2 liters.Rate this question:
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- 8.
Sebotol minuman berkarbonasi mengandung sejumlah asam benzoat/asam lemah (Ka C6H5COOH = 1.10–10). Jika konsentrasi minuman tersebut 0,4 M, maka pH minuman itu adalah ....
- A.
2-log 5
- B.
5,5 - log 2
- C.
3 - log 5,5
- D.
5 - log 2
- E.
8,5 + log 2
Correct Answer
B. 5,5 - log 2Explanation
The correct answer is 5,5 - log 2. This is because the pH of a solution can be calculated using the formula pH = -log[H+]. In this case, the concentration of the solution is given as 0.4 M. Since the concentration of H+ ions is not directly given, we need to calculate it using the Ka value of the weak acid present in the solution. However, the Ka value is not provided in the question. Therefore, we cannot determine the exact pH of the solution. The given answer, 5.5 - log 2, is not a valid explanation for the correct answer.Rate this question:
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- 9.
KOH tepat bereaksi dengan 25 mL larutan H2SO4 1 M (Ar K: 39, H : 1, O:16, S:32) : 2KOH + H2SO4 → K2SO4 + 2H2O Massa KOH yang bereaksi untuk menetralkan asam sulfat tersebut adalah ....
- A.
1,40 gram
- B.
4,90 gram
- C.
2,45 gram
- D.
2,80 gram
- E.
5,60 gram
Correct Answer
D. 2,80 gramExplanation
The balanced chemical equation shows that 2 moles of KOH react with 1 mole of H2SO4. Since the concentration of H2SO4 is given as 1 M, this means that there is 1 mole of H2SO4 in 1 liter of solution. Since we are using 25 mL of H2SO4 solution, the number of moles of H2SO4 is 1/1000 * 1 * 25 = 0.025 moles. According to the stoichiometry of the balanced equation, this means that we need 2 * 0.025 = 0.05 moles of KOH to react with the H2SO4. The molar mass of KOH is 39 + 16 + 1 = 56 g/mol, so the mass of KOH needed is 0.05 * 56 = 2.8 grams. Therefore, the correct answer is 2.80 gram.Rate this question:
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- 10.
Salah satu polutan penyebab hujan asam adalah gas SO3. Gas tersebut bereaksi dengan air membentuk asam Persamaan reaksi: SO3(g) + H2O( l) →H2SO4(g) Jika 16 gram gas SO3 (Ar H : 1, S : 32, dan O : 16), bereaksi dengan air, maka massa asam sulfat yang dihasilkan sebanyak ...
- A.
3,6 gram
- B.
16,0 gram
- C.
19,6 gram
- D.
80 gram
- E.
98 gram
Correct Answer
C. 19,6 gramExplanation
Gas SO3 bereaksi dengan air membentuk asam sulfat (H2SO4) sesuai dengan persamaan reaksi yang diberikan. Dalam persamaan tersebut, setiap 1 mol (80 gram) SO3 akan menghasilkan 1 mol (98 gram) H2SO4. Oleh karena itu, jika 16 gram SO3 bereaksi dengan air, maka massa asam sulfat yang dihasilkan dapat dihitung dengan menggunakan perbandingan mol. Berdasarkan perbandingan mol, massa asam sulfat yang dihasilkan adalah 16 gram SO3 x (98 gram H2SO4 / 80 gram SO3) = 19,6 gram H2SO4.Rate this question:
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Sekolah Global Mandiri - Wilayah Negara Kesatuan Republik Indonesia
Sekolah Global Mandiri - Wilayah Negara Kesatuan Republik Indonesia
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Seleksi Penerimaan Mahasiswa Baru Poltekkes Hermina Tahun 2020
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