Question

a) Calculate the pH of resulting solution if 25.0 mL of 0.250M Hcl(aq) is added to 15.0 mL of 0.350 M NaOH (aq)

b) Determine the pH of a solution when 22.8 mL of 0.026 M HNO3 is mixed with 15.8mL of 0.0090 M HCl

Answer 1

a)

Molarity of HCl = 0.250 M

Volume of HCl = 25.00 mL = 0.025 L

Moles of HCl = molarity x volume in Litres = 0.250 M X 0.0250 L = 0.00625 moles

Molarity of NaOH = 0.350 M

Volume of NaOH = 15.00 mL = 0.015 L

Moles of NaOH = molarity x volume in Litres = 0.350 M X 0.015 L = 0.00525 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 HCl + NaOH)

= (0.00625 moles - 0.00525 moles) /(0.025 L + 0.015 L)

= 0.025 M

[H+] = 0.025 M

pH = -log[H+] = - log [0.025] = 1.6

Therefore, pH = 1.6

b)

Moles of HNO3 = molarity x volume in Litres = 0.026 M X 0.0228 L = 0.0005928 moles

Moles of HNO3 = molarity x volume in Litres = 0.009 M X 0.0158 L = 0.0001422 moles

[H+] = moles of HNO3 + moles of HCl / volume of (HNO3+HCl) in Litres

       = 0.0005928 moles +0.0001422 moles / (0.0228 L + 0.0158 L)

      = 0.019 M

[H+]= 0.019 M

pH = - log [H+] = - log [0.019] = 1.72

Therefore,

pH = 1.72