Question

Find the pH of each mixture of acids.

a) 8.0×10−2 M in HNO3 and 0.175 M in HC7H5O2

b) 1.5×10⁻² M in HBr and 2.0×10⁻² M in HClO4

c) 9.0×10⁻² M in HF and 0.230 M in HC6H5O

d)0.100 M in formic acid and 4.5×10⁻² M in hypochlorous acid

Answer 1

Part a

[HNO3] = [H+] = 8*10^-2 mol/L

[HC7H5O2] = [H+] = 0.175 mol/L

After mixing up

Total moles of H+ = 8*10^-2 + 0.175 = 0.255 moles

Total volume = 1 + 1 = 2 L

New concentration [H+] = 0.255/2 = 0.1275 M

pH = - log [H+] = - log (0.1275) = 0.894

Part b

[HBr] = [H+] = 1.5*10^-2 mol/L

[HClO4] = [H+] = 2*10^-2 mol/L

After mixing up

Total moles of H+ = 1.5*10^-2 + 2*10^-2 = 0.035 moles

Total volume = 1 + 1 = 2 L

New concentration [H+] = 0.035/2 = 0.0175 M

pH = - log [H+] = - log (0.0175) = 1.76

Part c

[HF] = [H+] = 9*10^-2 mol/L

[HC6H5O] = [H+] = 0.230 mol/L

After mixing up

Total moles of H+ = 9*10^-2 + 0.230  = 0.320 moles

Total volume = 1 + 1 = 2 L

New concentration [H+] = 0.320/2 = 0.160 M

pH = - log [H+] = - log (0.160) = 0.796

Part d

hypochlorous acid = HClO

Formic acid = HCOOH

[HCOOH] = [H+] = 0.100 mol/L

[HClO] = [H+] = 4.5*10^-2 mol/L

After mixing up

Total moles of H+ = 0.100 + 4.5*10^-2 = 0.145 moles

Total volume = 1 + 1 = 2 L

New concentration [H+] = 0.145/2 = 0.0725 M

pH = - log [H+] = - log (0.0725) = 1.14