2.Calculate the pH of the solution after the addition of 3.53×10-2 moles of pota
ID: 1010190 • Letter: 2
Question
2.Calculate the pH of the solution after the addition of 3.53×10-2 moles of potassium hydroxide to 155 mL of this solution. (Assume that the volume does not change upon adding potassium hydroxide)pH =
3.When a 29.6 mL sample of a 0.327 M aqueous hydrofluoric acid solution is titrated with a 0.370 M aqueous sodium hydroxidesolution, what is the pH after 39.2 mL of sodium hydroxide have been added? pH =
4.A 48.0 mL sample of a 0.406 M aqueous hypochlorous acid solution is titrated with a 0.342 M aqueous sodium hydroxide solution. What is the pH after 19.1 mL of base have been added?pH =
Explanation / Answer
3.
Volume of HF = 29.6 ml = 0.0296 L
Concentration of HF =0.327 M
Concentration of NaOH =0.370 M
Volume of NaOH = 39.2 ml = 0.0392 L
Moles of HF = 0.0296 L x 0.327 M=0.0097
Moles of NaOH = 0.0392 L x 0.370 M=0.0145
moles OH- in excess = 0.0145- 0.0097 =0.0048
total volume = 0.0296 + 0.0392 = 0.0688
[OH-]= 0.0048/ 0.0688=0.0698 M
pOH = - log 0.0698 =1.156
pH = 14 - pOH = 14 – 1.156 = 12.844
The PH is = 12.44
4.
Volume of HClO = 48.0 ml = 0.048 L
Concentration of HClO =0.406 M
Concentration of NaOH =0.342 M
Volume of NaOH = 19.1 ml = 0.0191 L
HClO + NaOH NaClO + H2O
Moles of HClO = 0.048 L x 0.406 M=0.0195
Moles of NaOH = 0.0191 L x 0.342 M=0.0065
moles H+ in excess = 0.0195 - 0.0065 =0.013
total volume = 0.048L +0.0191 = 0.0671
[H+]= 0.013/ 0.0671=0.194M
The Ka of hypochlorous acid is 3.5× 10-8.
pH = ½ × pK – ½ × log [AH]
pK = log(3.5 × 10-8) = 7.456
pH = ½ × 7.456 ½ × log0.194
= 3.728 (0.712) = 4.44
The PH is 4.44
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