How many moles of KOH(s) must be added to 200ml of a .100 M solution of Hn03 to bring the PH to 2.00?

Question

how many moles of KOH(s) must be added to 200ml of a .100 M solution of Hn03 to bring the ph to 2.00?

Answer 1

pH = -log [H+]

pH = 2 => [H+] = 10^-2 = 0.01 M

HNO3 --> H+ + NO3-

At the final reaction :

[HNO3] = [H+] = 0.01 M

Mol of HNO3 = 0.2 L x 0.01 M = 0.002 mol

The initial of HNO3 = 0.2 L x 0.1 M = 0.02 mol

KOH + HNO3 --> KNO3 + H2O

ymol...0.02mol

ymol...ymol

----------------------------

0mol...0.002mol

Moles of KOH = y mol = 0.02 mol - 0.002 mol = 0.018 mol