Question

Calculate the pH of the following solution assuming ideal (2 sig. figures):

62.2 mL of 0.15 M HCl(aq) was added to 52.1 mL of 0.12 M NaOH(aq)

Answer 1

The final pH of the solution is 1.26.

Initial concentrations:

HCl: 0.15 M

NaOH: 0.12 M

Volumes:

HCl: 62.2 mL = 0.0622 L

NaOH: 52.1 mL = 0.0521 L

Initial moles:

HCl: 0.0622 L * 0.15 M = 0.00933 mol

NaOH: 0.0521 L * 0.12 M = 0.006252 mol

Limiting reactant:

NaOH is the limiting reactant because it has fewer moles (0.006252 mol) than HCl (0.00933 mol).

Excess reactant:

HCl is the excess reactant.

Excess moles:

HCl: 0.00933 mol - 0.006252 mol = 0.003078 mol

Final moles:

`H_3O^+` (from NaOH): 0.006252 mol

`Cl^-` (from HCl): 0.00933 mol

`OH^-` (from excess HCl): 0.003078 mol

Final concentration of H3O+:

[
`H_3O^+`] = 0.006252 mol / (0.0622 L + 0.0521 L) = 0.0533 M

pH:

pH = -log10[
`H_3O^+`]= -log10(0.0533) = 1.26

Question:-

Calculate the pH of the following solution assuming ideal (2 sig. figures):

62.2 mL of 0.15 M HCl(aq) was added to 52.1 mL of 0.12 M NaOH(aq)