Question

1. a 25.00 ml sample of h₃po₄(aq) requires 31.00 ml of 0.2320 m naoh for its titration to the second equivalence point. what is the molarity of the acid?
2. a 0.054 m hno₂ solution is titrated with a koh solution. what is the h⁺ half - way to the equivalence point?
3. a student titrates an unknown monoprotic acid with a naoh solution from a buret. after the addition of 12.35 ml of naoh, the ph of the solution is 5.22. the equivalence point is reached at 24.70 ml of naoh. what is the kₐ of the acid?

Explanation:

Step1: Write the balanced chemical equation for the titration of $H_3PO_4$ with $NaOH$ at the second - equivalence point

$H_3PO_4 + 2NaOH=Na_2HPO_4 + 2H_2O$. The mole ratio of $H_3PO_4$ to $NaOH$ is $n_{H_3PO_4}:n_{NaOH}=1:2$.

Step2: Calculate the number of moles of $NaOH$

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters. $V_{NaOH}=31.00\ mL = 0.03100\ L$ and $M_{NaOH}=0.2320\ M$. So, $n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.2320\ mol/L\times0.03100\ L = 0.007192\ mol$.

Step3: Calculate the number of moles of $H_3PO_4$

Since $n_{H_3PO_4}=\frac{1}{2}n_{NaOH}$, $n_{H_3PO_4}=\frac{1}{2}\times0.007192\ mol = 0.003596\ mol$.

Step4: Calculate the molarity of $H_3PO_4$

$V_{H_3PO_4}=25.00\ mL = 0.02500\ L$. Using the formula $M=\frac{n}{V}$, $M_{H_3PO_4}=\frac{n_{H_3PO_4}}{V_{H_3PO_4}}=\frac{0.003596\ mol}{0.02500\ L}=0.14384\ M\approx0.1438\ M$.

for question 11:
At the half - equivalence point of a weak acid - strong base titration, $[H^+]=K_a$. For $HNO_2$, a weak acid, at the half - equivalence point, $[H^+]=K_a$ of $HNO_2$. But we need to calculate it in terms of the given information. First, we know that for a weak acid $HA$ in water: $HA
ightleftharpoons H^++A^-$, and $K_a=\frac{[H^+][A^-]}{[HA]}$. At the half - equivalence point, $[HA]=[A^-]$, so $[H^+]=K_a$. We are not given the $K_a$ value of $HNO_2$ directly, but if we assume we know it (or look it up, $K_a$ of $HNO_2 = 4.5\times10^{-4}$), at the half - equivalence point $[H^+]=4.5\times10^{-4}\ M$.

for question 12:

Step1: Determine the moles of acid and base at different points

Let the molarity of the acid be $M_a$ and that of $NaOH$ be $M_b$. At the equivalence point, $n_a = n_b$, where $n_a$ is the moles of acid and $n_b$ is the moles of base. Let $M_b$ be the molarity of $NaOH$. We know that at the equivalence point, $V_b = 24.70\ mL$ and at a certain point $V_{b1}=12.35\ mL$. Since $V_{b1}=\frac{1}{2}V_b$ at this point, we are at the half - equivalence point.

Step2: Use the pH at the half - equivalence point to find $K_a$

At the half - equivalence point of a weak acid - strong base titration, $pH = pK_a$. Given $pH = 5.22$, then $pK_a=5.22$. Using the formula $K_a = 10^{-pK_a}$, $K_a=10^{- 5.22}=6.02\times10^{-6}$.

Answer:

$0.1438\ M$