Calculate the resulting pH when 100. mL of 0.200 M HOCl is mixed with

Question

Calculate the resulting pH when 100. mL of 0.200 M HOCl is mixed with

a. 75 mL of 0.15 M KOH

b. 275 mL of 0.15 M KOH

Answer 1

HOCl + KOH ==> H2O + KClO .. balanced equation

a) mols HOCl = 100. ml x 1 L / 1000 ml x 0.200 mol / L = 0.0200 mols

mols KOH = 75 ml x 1 L / 1000 ml x 0.15 mol / L = 0.01125 mols

HOCl + KOH ==> H2O + KClO

0.0200.........0.01125.......0............0...............Initial

-0.01125....-0.01125...+0.01125...+0.01125.....Change

8.75x10^-3........0..........0.01125....0.001125.....Equilibrium

This creates a buffer since we have a weak acid (HOCl) and the conjugate base (ClO^-). We will use the Henderson Hasselbalch equation to find the pH. Final volume of solution at this point is 100 ml + 75 ml = 175 ml = 0.175 L.

[HOCl] = 8.75x10^-3 mol / 0.175 L = 0.050 M

[ClO^-] = 0.001125 mol / 0.175 L = 6.43x10^-3 M

pH = pKa + log[ClO^-] / [HClO] and pKa for HOCl = 7.46 (this should be given in the problem. I looked it up)

pH = 7.46 + log (6.43x10^-3) /0.05 = 7.46 -0.89

pH = 6.57

(b) mols HOCl = 0.020 moles (see above)

mols KOH = 275 ml x 1 L / 1000 ml x 0.15 mol/L = 0.0413 mols

HOCl + KOH ==> H2O + KClO

0.0200.....0.0413.........................0................Initial

-0.0200....0.0200...................+0.0200..........Change

0...............0.0213,,,,,,,,,,,,,,,,,,,,00200...........Equilibrium

This is NOT a buffer, since all of the weak acid has been consumed and converted to KClO, with additional KOH left over (0.0213 mols). The pH will predominantly be determined by the strong base, KOH. Final volume at this point is 100 ml + 275 ml = 375 ml = 0.375 L

[KOH] = 0.0213 mols / 0.375 L = 0.0568 M

pOH = -log [OH-] = -log 00568

pOH = 1.25

pH = 14 - pOH

pH = 14 - 1.25

pH = 12.8