Question

A mixture is created containing 250 mL of 0.200 M KOH and 300 mL of 0.500 M H₂SO₄. What is the pH of the final solution?

Answer 1

we have:

Molarity of H+ = 2*[H2SO4] = 2*0.500 M = 1.00 M

Volume of H+ = 300 mL

Molarity of OH- = 0.200 M

Volume of OH- = 250 mL

mol of H+ = Molarity of H+ * Volume of H+

mol of H+ = 1 M * 300 mL = 300 mmol

mol of OH- = Molarity of OH- * Volume of OH-

mol of OH- = 0.2 M * 250 mL = 50 mmol

We have:

mol(H+) = 300 mmol

mol(OH-) = 50 mmol

50 mmol of both will react

remaining mol of H+ = 250 mmol

Total volume = 550 mL

[H+]= mol of acid remaining / volume

[H+] = 250 mmol/550.0 mL

= 0.4545 M

we have below equation to be used:

pH = -log [H+]

= -log (0.4545)

= 0.342

Answer: 0.342