The pH of the mixture of 800 ml of 0.1 M HCl and 200 ml of 0.3 M NaOH is close to :
1.70
This is a limiting-reagent neutralisation, so work in moles rather than concentrations until the end.
Step 1 — moles of acid. \(n_{H^{+}} = 0.800 \times 0.1 = 0.08\) mol.
Step 2 — moles of base. \(n_{OH^{-}} = 0.200 \times 0.3 = 0.06\) mol.
Step 3 — neutralise. H+ and OH- react 1 : 1, so 0.06 mol of each is consumed, leaving
\(0.08 - 0.06 = 0.02\) mol of excess H+. The acid is in excess, so the solution must be acidic — which already eliminates the basic and near-neutral options.
Step 4 — find the new concentration. Total volume is \(800 + 200 = 1000\) mL = 1.000 L, so
\([H^{+}] = \frac{0.02}{1.000} = 0.02\) M.
Step 5 — take the pH.
\(\mathrm{pH} = -\log(0.02) = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.30 = 1.70\).
A quick check on plausibility: the original acid at 0.1 M would have had pH 1.00, and partial neutralisation plus dilution must raise that figure somewhat — 1.70 fits, whereas 1.10 would imply almost no neutralisation occurred.
Note that dilution alone is accounted for by using the combined volume; forgetting it is the commonest source of error in this type of calculation.
Hence the pH is close to 1.70.
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