Question
The weak acid \(CH_3COOH\) has a \(pK_a\) of 4.76. A solution is prepared by mixing 500.mL of 0.150M \(CH_3COOH\)(aq) and 0.0200 mol of \(NaOH(s)\). Which of the following can be used to calculate the pH of the solution?
▶️Answer/Explanation
Ans:C
Using the volume and molarity of the weak acid, 0.0750mol of \(CH_3COOH\) was initially in solution. After adding 0.0200mol of \(NaOH\), the solution contains 0.0550mol of the weak acid and 0.0200mol of its conjugate-base produced from the neutralization. Therefore, \(pH=4.76+log(\frac{0.0200}{0.0550})=4.32\)
Question
Which of the following provides the correct mathematical expression to calculate the pH of a solution made by mixing 10.0ml of 1.00M \(HCl\) and 11.0mL of 1.00M \(NaOH\) at 25°C?
A \(pH=−log(0.0010)=3.00\)
B\( pH=14.00+log(0.0010)=11.00\)
C \(pH=14.00+log(\frac{0.0010}{0.0210})=12.68\)
D \(pH=14.00+log(\frac{0.0010}{0.0110})=12.96\)
▶️Answer/Explanation
Ans:C
From the volumes and molarities of the solutions, 0.0100mol of \(HCl\) was mixed with 0.0110mol \(NaOH\) in a total volume of 21.0mL(0.0210L). The HCl will be completely neutralized, leaving 0.0010molOH− ions in solution. Given that \(14.00=pH+pOH\) and \(pOH=−log[OH^−]\) , \(pH=14.00+log(\frac{0.0010}{0.0210})=12.68\)
Question
A buffer solution is formed by mixing equal volumes of 0.12M \(NH_3(aq)\) and 0.10M \(HCl(aq)\), which reduces the concentration of both solutions by one half. Based on the \(pK_a\) data given in the table, which of the following gives the pH of the buffer solution?
A \(pH=−log(0.050)=1.30\)
B \(pH=9.25+log(\frac{0.010}{0.050})=8.55\)
C \(pH=9.25+log(\frac{0.060}{0.050})=9.32\)
D \(pH=14.00−(−log(0.010))=12.00\)
▶️Answer/Explanation
Ans:B
\(HCl(aq)\) is a strong acid that reacts completely with the weak base \(NH_3(aq)\), producing 0.050M \(NH_4^+\)(aq). The molarity of the excess \(NH_3(aq)\) is 0.010M. The pH of the resulting buffer solution calculated using the Henderson-Hasselbalch equation is equal to 8.55.
Question
A solution is prepared by adding 100 mL of 1.0 M \(HC_{2}H_{3}O_{2}(aq)\) to 100 mL of 1.0 M \(NaC_{2}H_{3}O_{2}(aq)\). The solution is stirred and its pH is measured to be 4.73. After 3 drops of 1.0 M HCl are added to the solution, the pH of the solution is measured and is still 4.73. Which of the following equations represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH?
(A) \(H_2O^{+}(aq) + OH(aq) → 2 H_{2}O(l)\)
(B) \(H_3O^{+}(aq) + Cl ̄(aq) → HCl(g) + H_{2}O(l)\)
(C)\( H2O^{+}(aq) + C_{2}H_{3}O_{2}^{ ̄}(aq) → HC_{2}H_{2}O_{2}(aq) + H_{2}O(l)\)
(D) \(HgO^{+}(aq) + HC_{2}H_{3}O_{2}(aq)\) → \(H_{2}C_{2}H_{3}O_{2}^{+}(aq) + H_{2}O (l)\)
▶️Answer/Explanation
Ans:C
Question
Sodium hydroxide is a strong base. This means that __________.
A) aqueous solutions of NaOH contain equal concentrations of \(H^+\) (aq) and \(OH^-\) (aq)
B) NaOH does not dissociate at all when it is dissolved in water
C) NaOH dissociates completely to \(Na^+\)(aq) and \(OH^-\)(aq) when it dissolves in water
D) NaOH cannot be neutralized by a weak acid
E) NaOH cannot be neutralized by ordinary means
▶️Answer/Explanation
Ans: C