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pH = -log10(Ka) - log10([HA]/[A-])= 4.20 - log10([HA]/[A-])

in which

  • [HA] = 15.265(g) / 122.1(g.mol-1) / 500(mL) = 2.500*10-4 mol/mL

and

  • [A-] = 12.97(g) / 144.1(g.mol-1) / 500(mL) = 1.800*10-4 mol/mL

So pH = 4.20 - log10([2.5]/[1.8]) = 4.20 - log10(1.39) = 4.20 - 0.14 = 4.06

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Q: What is the pH of a solution that results from mixing 15.265 g of benzoic acid C6H5COOH MM equals 122.1 and 12.97g of sodium benzoate NaC6H5COO MM equals 144.1 in water to produce 500 ml of solution?
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