Find molarity of each solution

Solution A: $NaOH$

Given $4\,g$ of $NaOH$ dissolved to make $100\,L$ solution.

Molar mass of $NaOH = 40\,g\,mol^{-1}$

$\text{moles of } NaOH = \frac{4}{40} = 0.1\,mol$

Volume $= 100\,L$

$M_A = \frac{0.1}{100} = 10^{-3}\,M$

Since $NaOH$ is a strong base, $[OH^-] = 10^{-3}\,M$

Solution B: $H_2SO_4$

Given $9.8\,g$ of $H_2SO_4$ dissolved to make $100\,L$ solution.

Molar mass of $H_2SO_4 = 98\,g\,mol^{-1}$

$\text{moles of } H_2SO_4 = \frac{9.8}{98} = 0.1\,mol$

Volume $= 100\,L$

$M_B = \frac{0.1}{100} = 10^{-3}\,M$

Treating $H_2SO_4$ as a dibasic strong acid for this level, $[H^+] = 2 \times 10^{-3}\,M$

Calculate moles taken for mixing

From 40 L of solution A

$\text{moles of } OH^- = 10^{-3} \times 40 = 0.04\,mol$

From 10 L of solution B

$\text{moles of } H_2SO_4 = 10^{-3} \times 10 = 0.01\,mol$

Since each mole of $H_2SO_4$ gives $2$ moles of $H^+$, $\text{moles of } H^+ = 2 \times 0.01 = 0.02\,mol$

Neutralization reaction

$H^+ + OH^- \rightarrow H_2O$

Initially:

• $OH^- = 0.04\,mol$
• $H^+ = 0.02\,mol$

After neutralization, excess $OH^-$: $0.04 - 0.02 = 0.02\,mol$

Find final concentration of excess $OH^-$

Total volume after mixing: $40 + 10 = 50\,L$

$[OH^-] = \frac{0.02}{50} = 4 \times 10^{-4}\,M$

Calculate pOH and pH

$pOH = -\log(4 \times 10^{-4})$

$pOH = 4 - \log 4$

Using $\log 4 \approx 0.6$, $pOH \approx 4 - 0.6 = 3.4$

$pH = 14 - 3.4 = 10.6$

Final answer

$\boxed{10.6}$

Comparison with stored correct answer

Stored correct answer: $10.6$

My derived answer matches the stored answer.