Ans 1 assume 1 L of solution
[Acid] = 0.85 M;
[salt] = 0.75 M
pKa = – log{(Ka) / M} = – log (1.8 x 10-4) = 3.75
pH = pKa + log[salt] / [acid]) = 3.75 + log(0.75M / 0.85M)
pH = 3.75 + log(0.88) = 3.75 - 0.056 = 3.69
Ans 2 -
assume 1 L of solution
Add 8 g NaOH
Moles of NaOH added = 8/40 = 0.2 moles
The concentration will change
[Acid] = 0.85 - 0.2 = 0.65 M;
[salt] = 0.75 + 0.2 = 0.95 M
pKa = – log{(Ka) / M} = – log (1.8 x 10-4) = 3.75
pH = pKa + log[salt] / [acid]) = 3.75 + log(0.95M / 0.65M)
pH = 3.75 + log(1.46) = 3.75 + 0.16 = 3.91
Ans 3
assume 1 L of solution
Add 16 g HBr
Moles of HBr added = 16/80.91 = 0.2 moles
The concentration will change
[Acid] = 0.85 + 0.2 = 1.05 M;
[salt] = 0.75 - 0.2 = 0.55 M
pKa = – log{(Ka) / M} = – log (1.8 x 10-4) = 3.75
pH = pKa + log[salt] / [acid]) = 3.75 + log(0.55M / 1.05M)
pH = 3.75 + log(0.52) = 3.75 - 0.28 = 3.47
Ans 4
assume 1 L of solution
Add 60.7 g HBr
Moles of HBr added = 60.7/80.91 = 0.75 moles
The concentration will change
[Acid] = 0.85 + 0.75 = 1.60 M;
[salt] = 0.75 - 0.75 = 0 M
pKa = – log{(Ka) / M} = – log (1.8 x 10-4) = 3.75
pH = pKa + log[salt] / [acid]) = 3.75 + log(0M / 1.60M)
pH = 3.75 + log(0) = 3.75 = 3.75
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