Question

Calculate the hypothetical pH AFTER addition of 10.00 mL of reagent for the titration of 50.00 mL of 0.0800 M HCl with 0.1000 M NaOH at 25°C.

Answer 1

Volume of HCl = 50 ml = 0.050 L

Moles of HCl present = molarity x volume

= 0.0800 M * 0.050 L

= 0.004 moles

Molarity of NaOH = 0.100 M

Volume of NaOH = 10.00 ml = 0.010 L

Moles of NaOH added = 0.100 M*0.010 L = 0.001 mol

Moles of HCl remaining = 0.004 - 0.001 = 0.003 mol

Moles of H⁺ = 0.003 mol

Total volume = 60 ml = 0.060 L

[H⁺] = 0.003 mol / 0.060 L = 0.05 M

pH = -log[H⁺]

= -log(0.05)

= 1.30