A 25.00-mL solution of 0.200 M HN03 was added to 30.00 mL of a KOH solution of i

Question

A 25.00-mL solution of 0.200 M HN03 was added to 30.00 mL of a KOH solution of inknown concentration. At 25 °C, the pH of the mixture was 11.350. If the solutions were ideal, what was the original concentration of the KOH? 2. An unknown amount of weak acid, HA, was dissolved in water to make 50.00 mL of solution and titrated with 0.0800 M NaOH. Data from a pH-meter showed an equivalence point upon addition of 32.80 mL of the base. When only 26.00 mL was added, the pH was 5.300. Calculate pKa 3. A 25.00-mL solution of 0.100 M HAc (Ka = 1.76 x 10-5) is titrated with 0.100 M NaOH. What is the pH of the solution when the volume of added NaOH is 3.00 mL in excess of that needed to reach the equivalence point? 4. A 20.0-mL solution of 0.120 M citric acid was titrated with 0.420 M KOH to the equivalence point. What volume of KOH is required to reach the equivalence point? Determine the pH at the equivalence point.

Explanation / Answer

Q1.

V = 25 mL of acid, M = 0.2 HNO3

V = 30 mL of KOH, M = X

if pH = 11.35

find original KOH

first

the reaction

HNO3 + KOH = KNO3 + H2O

mmol of acid = MV = 0.2*25 = 5 mmol

mmol of base = 5 + x

since we need 5 to neutralize, and "x" is the pH shown as 11.35

pOH = 14-pH = 14-11.35 = 2.65

[OH-] = 10^-pOH = 10^-2.65 = 0.002238 M

Vtotal = 25+30 = 55 mL

mmol of OH- = MV = 55*0.002238 = 0.12309 mmol

then..

mmol of OH- = 5+0.12309 = 5.12309

of which

[KOH] = 5.12309 / V = 5.12309/30 = 0.17076 M

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