A) Determine the pH during the titration of 27.0 mL of 0.158 M hydrobromic acid

Question

A) Determine the pH during the titration of 27.0 mL of 0.158 M hydrobromic acid by 0.201 M sodium hydroxide at the following points:

(1) Before the addition of any sodium hydroxide

(2) After the addition of 10.6 mL of sodium hydroxide

(3) At the equivalence point

(4) After adding 26.3 mL of sodium hydroxide

B) Determine the pH during the titration of 12.3 mL of 0.398 M hydrobromic acid by 0.333 M potassium hydroxide at the following points:

(1) Before the addition of any potassium hydroxide

(2) After the addition of 7.35 mL of potassium hydroxide

(3) At the equivalence point

(4) After adding 18.7 mL of potassium hydroxide

Please make sure its correct :) Thank you

Explanation / Answer

1 ) before addition of NaOH

HBr = 0.158 M

[H+] = 0.158 M

pH = -log [H+] = -log (0.158)

pH = 0.80

2) After the addition of 10.6 mL of sodium hydroxide

   millimoles of acid = 27 x 0.158 = 4.266

millimoles of base = 0.201 x 10.6 = 2.131

acid millimoles > base millimoles

[H+] = (4.266 - 2.131 ) / (27 +10.6) = 0.5678 M

pH = 1.25

3 ) at equivalence point :

HBr and NaOH both are strong . so

pH = 7

4) After adding 26.3 mL of sodium hydroxide

base millimoles = 26.3 x 0.201 = 5.286

acid millimoles = 27 x 0.158 = 4.266

base > acid

[OH-] =( 5.286 - 4.266 ) / (26.3 +27)

          = 0.0191 M

pOH = -log [OH-]

pOH = -log (0.0191)

pOH = 1.72

pH + pOH = 14

pH = 12.28

note : similarly sove B )problems

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