Question

Calculate the pH of a solution prepared by mixing 20.0 mL of 1.00M hydrochloric acid with 10.0 mL of 1.00M sodium hydroxide. (Indicate the limiting reactant and reactant in excess below)

please help me with this question step by step, is there any formula to solve this question?

Answer 1

HCl (aq.) + NaOH (aq.) ------------> NaCl (aq.) + H₂O (l)

Moles of HCl = 1.00 * 20.0 / 1000 = 0.0200 mol

Moles of NaOH = 1.00 * 10.0 /1000 = 0.0100 mol

According to the balanced equation provided above, 1 mol HCl requires 1 mol NaOH

So, NaOH is limiting reagent. Because 0.0100 < 0.0200

Remaining, moles of NaOH = 0.0200 - 0.0100 = 0.0100 mol

So, new concentration of NaOH = 0.0100*1000 / 30.0 = 0.333 M

[NaOH] = [OH-] = 0.333 M

pOH = - Log[OH-]

pOH = - Log(0.333)

pOH = 0.478

We know that,

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 0.478

pH = 13.5