What is the molarity of the acetic acid solution if 31.7 mL of a 0.235 M KOH sol

Question

What is the molarity of the acetic acid solution if 31.7 mL of a 0.235 M KOH solution is required to titrate 26.0 mL of a solution of HC2H3O2?

HC2H3O2(aq)+ KOH(aq) H2O(l)+ KC2H3O2(aq)

Express your answer with the appropriate units

If 33.0 mL of a 0.164 M NaOH solution is required to titrate 25.5 mL of a solution of H2SO4, what is the molarity of the H2SO4 solution?

H2SO4(aq)+2NaOH(aq)2H2O(l)+Na2SO4(aq)

Express your answer with the appropriate units.

A solution of 0.314 M KOH is used to titrate 15.5 mL of a 0.182 M H3PO4 solution. What volume, in milliliters, of the KOH solution is required?

H3PO4(aq)+3KOH(aq)3H2O(l)+K3PO4(aq)

Express your answer with the appropriate units.

please explain steps on how to solve the answer

Explanation / Answer

A)

mmol of base = M*V = 31.7*0.235 = 7.4495 mmol of OH-

ratio is !:! so, 7.4495 mol of acid are needed

M = mmol/ml = 7.4495/26 = 0.28651 mmol of acid

B)

calculat emmol of base

mmol = M*V = 33*0.164 = 5.412 mmol of base

then we need 5.412/2 = 2.706 mmol of acid

M = mmol/V = 2.706/25.5 = 0.106117

C)

calculate mmol of acid

M*V = 15.5*0.182 = 2.821 mmol of acid

ratio is 1:3 so

2.821*3 = 8.463 mmol of base is needed

then

V= mmol/M = 8.463/0.314 = 26.952 ml

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