Question

part b application questions: answer the following questions. show all work for calculations and include units and the correct number of significant figures. 1. what is the concentration of an arsenic acid solution if it required 9.84 ml 0.1083 m naoh to reach the equivalence point when 25.00 ml of the h₃aso₄ solution was titrated? the titration reaction is given by h₃aso₄ (aq) + 3 naoh (aq) → na₃aso₄ (aq) + 3 h₂o (l) 2. what is the molar mass of a diprotic acid if 0.1063 g of the acid required 15.47 ml of a 0.1164 m naoh solution to reach the equivalence point?

Explanation:

Step1: Calculate moles of NaOH for first - question

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume in liters. For the first question, $V_{NaOH}=9.84\ mL = 9.84\times10^{- 3}\ L$ and $M_{NaOH}=0.1083\ M$. So, $n_{NaOH}=0.1083\ mol/L\times9.84\times10^{-3}\ L = 0.1083\times9.84\times10^{-3}\ mol=1.065672\times10^{-3}\ mol$.

Step2: Determine moles of $H_3AsO_4$

From the balanced chemical equation $H_3AsO_4(aq)+3NaOH(aq)\to Na_3AsO_4(aq) + 3H_2O(l)$, the mole - ratio of $H_3AsO_4$ to $NaOH$ is $\frac{1}{3}$. So, $n_{H_3AsO_4}=\frac{1}{3}n_{NaOH}=\frac{1.065672\times10^{-3}\ mol}{3}=3.55224\times10^{-4}\ mol$.

Step3: Calculate concentration of $H_3AsO_4$

Use the formula $M=\frac{n}{V}$, where $V_{H_3AsO_4}=25.00\ mL = 25.00\times10^{-3}\ L$. So, $M_{H_3AsO_4}=\frac{3.55224\times10^{-4}\ mol}{25.00\times10^{-3}\ L}=0.01420896\ M\approx0.01421\ M$.

Step4: Calculate moles of NaOH for second - question

For the second question, $V_{NaOH}=15.47\ mL = 15.47\times10^{-3}\ L$ and $M_{NaOH}=0.1164\ M$. So, $n_{NaOH}=0.1164\ mol/L\times15.47\times10^{-3}\ L=1.791708\times10^{-3}\ mol$.

Step5: Determine moles of diprotic acid

For a diprotic acid ($H_2A$) reacting with $NaOH$ ($H_2A + 2NaOH\to Na_2A+2H_2O$), the mole - ratio of the diprotic acid to $NaOH$ is $\frac{1}{2}$. So, $n_{acid}=\frac{1}{2}n_{NaOH}=\frac{1.791708\times10^{-3}\ mol}{2}=8.95854\times10^{-4}\ mol$.

Step6: Calculate molar mass of diprotic acid

Use the formula $M=\frac{m}{n}$, where $m = 0.1063\ g$ and $n = 8.95854\times10^{-4}\ mol$. So, $M=\frac{0.1063\ g}{8.95854\times10^{-4}\ mol}=118.669\ g/mol\approx118.7\ g/mol$.

Answer:

1. The concentration of the $H_3AsO_4$ solution is $0.01421\ M$.
2. The molar mass of the diprotic acid is $118.7\ g/mol$.