Question

Titration of a Weak Monoprotic Acid with a Strong Base Volume of Base (mL) 1. Using the graph, determine the K, of the weak acid. 2. Suppose 100mL of the monoprotic acid referred to in the graph was titrated with 1M NaOH, determine the molarity of the weak acid. The corresponding balanced chemical equation is shown below. HA + NaOH + H2O + A + Nat 3. If the weak acid above was prepared by using 5g of the weak acid to make a 500ml aqueous solution, determine the molar mass of the weak acid.

Answer 1

1) First find out equivalence point on the graph.

From equivalence point, we get half equivalence point.

pH corresponding to half equivalence point is the pKa i e pH at half equivalence point = pKa = 4

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We have , pKa of weak acid (HA) = 4

We have relation, pKa = - log K a

Therefore, K a = 10 ^-pKa = 10 ^{- 4}

ANSWER : Ka of weak acid (HA) = 1 $\times$ 10 ^{- 4}

2) Consider a reaction, HA+ NaOH NaA + H₂O

From reaction, stoichiometric ratio = No of moles of acid / No of moles of base

stoichiometric ratio = No. of moles of HA / No. of moles of NaOH = 1/1=1

We have correlation, M _acid V _acid = M _base V _base stoichiometric ratio.

We have, V _acid = 100 ml , V _base = 10.0 ml , M _base = 1 M

$\therefore$ M _acid 100 ml = 1 M 10.0 ml 1

M _acid = 1 M 10.0 ml 1 / 100 ml =0.1 M

ANSWER : Molarity of weak acid (HA) = 0.1 M

C) We know that, Molarity = No. of moles of solute / Volume of solution in L

0.1 M weak acid HA is prepared by dissolving 5 g of HA in 500 ml of water.

Therefore,0.1 mol / L = No. of moles of HA / 0.500 L

Therefore, No. of moles of HA = ( 0.1 mol / L) 0.500 L = 0.05 mol

We have, No. of moles = Mass / Molar mass

Molar mass of HA = Mass / No. of moles = 5 g / 0.05 = 100 g /mol

ANSWER : Molar mass of weak acid HA = 100 g/mol