A 20.00 mL sample of a solution of a weak monoprotic acid HA is titrated with a

Question

A 20.00 mL sample of a solution of a weak monoprotic acid HA is titrated with a 0.2018 M solution of potassium hydroxide (KOH), a strong soluble base. After the addition of 17.27 mL of the potassium hydroxide solution the end point of the titration is reached.

a) Give the balanced equation for the above titration reaction.

b) Using the information given above, find the concentration of HA in the weak acid solution.

c) Will the pH at the equivalence point for the above titration be higher than 7.0, approximately equal to 7.0, or lower than 7.0? Justify your answer.

Explanation / Answer

A) .The balanced equation for this titration reaction will be:

HA + KOH = KA + H2O

B) the concentration of acid can be found out using the formula

M1V1 = M2V2

Where M and V are the molarity and volumes of acid and base.

So putting the values in the formula,

M1 × 20 = 0.2018 × 17.27

M1 = 0.1742 M

So the concentration of weak avid HA is 0.1742 mol/L

C) The pH at the equivalence point will be higher than 7 because due to the titration of weak acid by the strong base , the conjugate base of the weak acid is formed in the solution. This will cause the pH of solution at the equivalence point more than 7 , i.e basic.

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