20 mL of $0.1 ~\mathrm{M} ~\mathrm{NaOH}$ is added to $50 \mathrm{~mL}$ of $0.1 ~\mathrm{M}$ acetic acid solution. The $\mathrm{pH}$ of the resulting solution is ___________ $\times 10^{-2}$ (Nearest integer)

Given : $\mathrm{pKa}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76$

$\log 2=0.30$

$\log 3=0.48$

First, we need to find the moles of NaOH and acetic acid (CH₃COOH) in the solution: <br/><br/> Moles of NaOH = Volume × Molarity = 20 mL × 0.1 M = 2 mmol<br/><br/> Moles of acetic acid = Volume × Molarity = 50 mL × 0.1 M = 5 mmol <br/><br/> Since NaOH is a strong base, it will react with acetic acid to form acetate ions (CH₃COO⁻) and water: <br/><br/> CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O <br/><br/> 2 mmol of NaOH will react with 2 mmol of acetic acid, resulting in 2 mmol of acetate ions and leaving 3 mmol of acetic acid unreacted. <br/><br/> Next, we need to find the concentrations of acetic acid and acetate ions in the resulting 70 mL solution: <br/><br/> Concentration of acetic acid = Moles / Total volume = 3 mmol / 70 mL = 0.04286 M<br/><br/> Concentration of acetate ions = Moles / Total volume = 2 mmol / 70 mL = 0.02857 M <br/><br/> Now we can use the Henderson-Hasselbalch equation to find the pH of the resulting solution: <br/><br/> pH = pKa + log ([A⁻] / [HA]) <br/><br/> Given the pKa of acetic acid is 4.76, we can substitute the values: <br/><br/> pH = 4.76 + log (0.02857 / 0.04286) <br/><br/> Using the given log values, we can approximate the log value: <br/><br/> log (0.02857 / 0.04286) ≈ log (2/3) ≈ log 2 - log 3 ≈ 0.30 - 0.48 = -0.18 <br/><br/> Now substitute this value back into the Henderson-Hasselbalch equation: <br/><br/> pH = 4.76 - 0.18 = 4.58 <br/><br/> To express the pH as the nearest integer multiplied by 10⁻², multiply the pH value by 100: <br/><br/> 4.58 × 100 = 458 <br/><br/> So, the pH of the resulting solution is approximately 458 × 10⁻².