Question

What is the pH of a solution prepared by mixing 45.67 mL of 0.8949 M HCl and 31.31 mL of 0.8566 M NaOH?

Answer 1

HCl:

Molarity = 0.8949 M

volume = 45.67 mL = 0.04567 L

Moles of HCl = Molarity x volume = 0.8949 M X 0.04567 = 0.041 moles

NaOH:

Molarity = 0.8566 M

volume = 31.31 mL = 0.03131 L

Moles of NaOH = Molarity x volume = 0.8566 M X 0.03131 L = 0.027 moles

But moles of HCl = 0.041 moles

Therefore, concentration of HCl is more than concentration of NaOH.

Hence at equivalence point, concentration of H+ is higher than concentration of OH- ions.

[H+] = moles of HCl - moles of NaOH / total volume (volume of NaOH+ HCl)

= (0.041 moles - 0.027 moles) /(0.04567 L + 0.03131 L)

= 0.1818 M

[H+] = 0.1818 M

pH = -log[H+]

= -log( 0.1818)

= 0.74

pH = 0.74

Therefore, pH = 0.74