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pH of mixture of three samples

Short answer

Combine hydrogen-ion amounts from all three samples: final H+ = (700e-6 + 300e-7 + 500e-8)/1500 = 4.90e-7 mol/l. Taking the negative log gives pH = 6.31.

700 ml of water sample A with pH of 6 is mixed with 300 ml of sample B with pH 7 and 500 ml of sample C with pH 8. Calculate the pH of the mixture of A, B and C.

The moles of H+^+ in each sample add on mixing, so the final hydrogen-ion concentration is the volume-weighted average. Buffering is ignored.

SampleVolume (ml)pH[H+][H^+] (mol/l)V×[H+]V \times [H^+] (ml⋅\cdotmol/l)
A700610−610^{-6}7.0×10−47.0\times10^{-4}
B300710−710^{-7}3.0×10−53.0\times10^{-5}
C500810−810^{-8}5.0×10−65.0\times10^{-6}
[H+]mix=7.0×10−4+3.0×10−5+5.0×10−6700+300+500=7.35×10−41500=4.90×10−7 mol/l\begin{aligned} [H^+]_{mix} &= \frac{7.0\times10^{-4} + 3.0\times10^{-5} + 5.0\times10^{-6}}{700+300+500} \\ &= \frac{7.35\times10^{-4}}{1500} = 4.90\times10^{-7}\ \text{mol/l} \end{aligned} pH=−log⁡10(4.90×10−7)=6.31pH = -\log_{10}(4.90\times10^{-7}) = 6.31

Answer: pH of the mixture = 6.31.

✓ Answer checked by Claude Opus 5.5 (high effort)

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