A nitric acid solution, HNO_3, will be titrated with potassium hydroxide, KOH. W

Question

A nitric acid solution, HNO_3, will be titrated with potassium hydroxide, KOH. What is the pH of the solution after adding 102.6 mL of 0.095 M KOH to the 65 mL of 0.15 M HNO_3 solution? What is the pH of the solution after adding 105.5 mL of 0.095 M KOH to the 65 mL of 0.15 M HNO_3 solution? Part 1: pH = 7.00 (Equivalence Point); Part 2: pH = 2.79 Part 1: pH = 5.11; Part 2: pH = 6.73 Part 1: pH = 6.05; Part 2: pH = 12.07 Part 1: pH = 7.00 (Equivalence Point); Part 2: pH = 11.20 Part 1; pH = 7.00 (Equivalence Point); Part 2: pH = 9.62

Explanation / Answer

both HNO3 and KOH are strong. They ionize completely

HNO3+KOH---->KNO3+H2O, from the reaction, 1 mole of KOH reacts with 1 mole of HNO3

moles of KOH in 102.6ml of 0.095M KOH= molarity* Volume (L) = 0.095*102.6/1000 = 0.00975

moles of HNO3 in 65ml of of 0.15M= 0.15*65/1000 = 0.00975 moles (this remained same in part-1 )

In the present case, excess is HNO3, excess amount = 0.00975-0.00975=0

Volume after mixing = 102.6+ 65= 167.6 ml = 167.6/1000L=0.1676 L

Since the acid and base neutralize and both or stonrg, the pH= 7.

part-2 : moles of KOH in 105.5ml of 0.095M = 0.095*105.5/1000=0.010032

moles of HNO3= 0.00975

Excess is KOH and excess by =0.010032-0.00975=0.000282

Volume after mixing = 105.5+65=170.5ml =170.5/1000 =0.1705

Concentratino of KOH= 0.000282/0.1705=0.001654

pOH= -log (0.001654)=2.78

pH= 14-2.78= 11.22 ( D is correct answer).

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