I don’t know how to do this question **Problem:**A 100.00 mL solution of 0.80 M in \( \text{HC}_6\text{H}_6\text{O}_6 \) is titrated with 0.80 M NaOH. Find the pH after 100.00 mL of NaOH has been added. For \( \text{HC}_6\text{H}_6\text{O}_6 \), \( K_a = 7.9 \times 10^{-5} \).---**Description:**This problem involves a titration between a weak acid (\( \text{HC}_6\text{H}_6\text{O}_6 \)) and a strong base (NaOH). You are asked to find the pH of the solution after an equivalent amount of NaOH has been added to the initial acid solution. **Steps to Solve:**1. **Initial Moles of Acid and Base:** - Calculate moles of \( \text{HC}_6\text{H}_6\text{O}_6 \): \( 0.100 \, \text{L} \times 0.80 \, \text{M} = 0.080 \, \text{mol} \). - Calculate moles of NaOH: \( 0.100 \, \text{L} \times 0.80 \, \text{M} = 0.080 \, \text{mol} \).2. **Determine the Reaction Progress:** - Since moles of acid and base are equal, they completely neutralize each other to form the conjugate base.3. **Calculate the Concentration of the Conjugate Base:** - The total volume after mixing = \( 100.00 \, \text{mL} + 100.00 \, \text{mL} = 200.00 \, \text{mL} = 0.200 \, \text{L} \). - Concentration of conjugate base \( (\text{C}_6\text{H}_5\text{O}_6^-) \) = \( \frac{0.080 \, \text{mol}}{0.200 \, \text{L}} = 0.40 \, \text{M} \).4. **Use

Consider the given information is as follows; Volume of HC6H6O6 = 100 .00 mL = 0.1 L Concentration of HC6H6O6 = 0.80 M = 0.80 mol/L Volume of NaOH = 100 .00 mL = 0.1 L Concentration of NaOH = 0.80 M = 0.80 mol/L Ka for HC6H6O6 = 7.9×10-5Determination of initial moles of HC6H6O6 and NaOH from the given volume and concentration is as follows; Moles of HC6H6O6=Concentration of HC6H6O × Volume of HC6H6O=0.80 mol/L × 0.1 L=0.08 molMoles of NaOH=Concentration of NaOH × Volume of NaOH=0.80 mol/L × 0.1 L=0.08 mol Total volume = 0.1 + 0.1 = 0.2 L Determination of pka from the given Ka is as follows; pKa=-log Ka=-log 7.9×10-5=-log 7.9 + log 10-5=-0.8976 - 5=4.10 Consider the reaction between HC6H6O6 and NaOH is as follows; Reaction: HC6H6O6 + NaOH → NaC6H6O6 + H2OInitial moles: 0.08 mol 0.08 mol 0 Equilibrium moles: 0.08-0.08 0.08-0.08 0.08 mol =0.0 mol =0.0 molEquilibrium conc. : 0.0 mol0.2 L=0 0.08 mol0.2 L=0.4 M Now, substitute the values in the formula given below as; pH = pKa + log NaC6H6O6HC6H6O6=4.10 + log 0.4 M0.0 M=4.10 + log 0=4.10 Hence, the pH of the given solution will be 4.10