Question

Determine the pH of a solution when 26.3 mL of 0.027 M HNO3 is mixed with 25.0 mL of

A) 0.0050 M NaOH.

B) distilled water.

C) 0.0150 M HCl.

D) 0.045 M KOH.

Please show all work.

Answer 1

Concentration of HNO3 = 0.027 M = 0.027 mol/L

Volume of HNO3 solution = 26.3 ml = 26.3 L / 1000 = 0.0263 L

Number of moles of HNO3 = 0.027 mol/L * 0.0263 L = 0.00071 mol

A) Volume of NaOH solution = 25.0 ml = 25.0 L / 1000 = 0.025 L

Concentration of NaOH = 0.0050 M = 0.0050 mol/L

Number of moles of NaOH = 0.025 L * 0.0050 mol/L = 0.000125 mol

Reaction :-. NaOH(aq) + HNO3(aq) ----> NaNO3(aq) + H2O (l)

From reaction, 1.0 mol of NaOH reacts with 1.0 mol of HNO3 then 0.000125 mol of NaOH will react with 0.000125 mol of HNO3.

Number of moles of HNO3 remaining = ( 0.00071 - 0.000125) mol = 0.000585 mol

Total volume of solution = (0.0263 + 0.025) L = 0.0513 L

Since HNO3 is a strong monoprotic acid hence

[H^+] = [HNO3] = 0.000585 mol / 0.0513 L = 0.0114 mol/L

pH = - log[H^+] = - log 0.0114 = - ( - 1.94) = 1.94

pH = 1.94

B) Volume of distilled water mixed = 25.0 ml = 25.0 L / 1000 = 0.025 L

Total volume of solution = ( 0.025 + 0.0263) L = 0.0513 L

Similarly [H^+] = [HNO3] = 0.00071 mol / 0.0513 L = 0.01384 mol/L

pH = - log 0.01384 = - ( - 1.86) = 1.86

pH = 1.86

C) Concentration of HCl = 0.0150 M = 0.0150 mol/L

Volume of HCl solution = 0.025 L

Number of moles of HCl = 0.0150 mol/L * 0.025 L = 0.000375 mol

Total number of moles of monoprotic acid = (0.000375 + 0.00071) mol = 0.001085 mol

Total volume of solution = (0.025 + 0.0263) L = 0.0513 L

Similarly, [H^+] = 0.001085 mol / 0.0513 L = 0.02115 mol/L

pH = - log 0.02115 = - ( - 1.67) = 1.67

pH = 1.67

D) Concentration of KOH = 0.045 M = 0.045 mol/L

Number of moles of KOH = 0.045 mol/L * 0.025 L = 0.001125 mol

Reaction :-. HNO3(aq) + KOH(aq) -----> KNO3(aq) + H2O (l)

From reaction, 1.0 mol of HNO3 reacts with 1.0 mol of KOH then 0.00071 mol of HNO3 will react with 0.00071 mol of KOH.

Number of moles of KOH remaining = (0.001125 - 0.00071) mol = 0.000415 mol

Total volume of solution after mixing = 0.0513 L

Since KOH is a strong monobasic hence

[OH^-] = [KOH] = 0.000415 mol / 0.0513 L = 0.00809 mol/L

pOH = - log [OH^-] = - log 0.00809 = - (-2.09) = 2.09

pH = 14.00 - pOH = 14.00 - 2.09 = 11.91

pH = 11.91