Question

1. When 50.00 mL of 0.100 M NaOH reacts with 25.00 mL of 0.300 M HX,...

1. When 50.00 mL of 0.100 M NaOH reacts with 25.00 mL of 0.300 M HX, a weak monoprotic acid, the pH of the resulting solution is 3.74. Calculate the Ka of the acid.

2. Will the equivalence point of ammonia plus HCl be higher, lower, or equal to 7? Explain.

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Homework Answers

Answer #1

1) [NaOH ] = molarity x volume in Litres = 0.1 M x 0.05 L = 0.005 mol

[HX] = 0.3 M x 0.025 L = 0.0075 mol

HX + NaOH ---------------> NaX + H2O

0.0075 mol 0.005 mol 0

-----------------------------------------------------------------------------

0.0075 - 0.005 0 0.005 mol

= 0.0025 mol

Hence,

[HX] = 0.0025 mol

[NaX] = 0.005 mol

pH = -logKa + log [NaX] /[HX]

3.74 = -logKa + log [0.005]/[0.0025]

logKa = log [0.005]/[0.0025] - 3.74

Hence,

Ka = 0.000364

Therefore,

Ka of the acid =  0.000364

2) Equivalence point of ammonia plus HCl

  It will be lower than 7.

Because, HCl is a strong acid and ammonia is weak base.

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