Question

(3)   A 423 mL sample of 1.142 M HCl is mixed with 509 mL sample of NaOH (which has a pH of 13.30). What is the pH of the resulting solution? (Give answer to two decimal places)                  

Answer 1

HCl:

volume = 423 mL = 0.423 L

molarity = 1.142 M

moles = molarity x volume = 1.142 M x 0.423 L = 0.48 mol

NaOH:

volume = 509 mL = 0.509 L

pH = 13.3

pOH = 14 - pH = 14 - 13.3 = 0.7

[OH-] = 10^-pOH = 10^-0.7 =0.2 M

molarity of NaOH = 0.2 M

moles of NaOH = molarity x volume = 0.2 M x 0.509 L = 0.102 mol

So, moles of HCl is more than moles of NaOH.

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 HCl + NaOH)

= ( 0.48 moles -0.102 moles) /(0.423 L + 0.509 L)

= 0.4 M

[H+] = 0.4 M

pH = -log[H+]

= -log( 0.4)

= 0.397

pH = 0.397

Hence, pH of the resulting solution = 0.397