Question

A chemistry graduate student is given 125.mL of a 1.40M chloroacetic acid HCH2ClCO2 solution. Chloroacetic acid is a weak acid with =Ka×1.310−3. What mass of KCH2ClCO2 should the student dissolve in the HCH2ClCO2 solution to turn it into a buffer with pH =2.97? You may assume that the volume of the solution doesn't change when the KCH2ClCO2 is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Answer 1

We can solve given problem in following steps.

Step 1 : Calculation of ratio of [ ClCH2COOK ] / [ ClCH2COOH ]

We have Henderson's equation, pH = pKa + log [ Salt ] / [ Acid ]

$\therefore$ pH = - log Ka + log [ ClCH2COOK ] / [ ClCH2COOH ]

$\therefore$ 2.97 = - log ( 1.3 $\times$ 10 -03 ) + log [ ClCH2COOK ] / [ ClCH2COOH ]

2.97 =2.89 + log [ ClCH2COOK ] / [ ClCH2COOH ]

log [ ClCH2COOK ] / [ ClCH2COOH ] = 2.97 - 2.89 = 0.08

Taking antilog on both sides, we get   [ ClCH2COOK ] / [ ClCH2COOH ] = 1.20

[ ClCH2COOK ] = 1.20 $\times$ [ ClCH2COOH ]

We can write , No. of moles of ClCH2COOK = 1.20 $\times$ No. of moles of ClCH2COOH

Step 2 : Calculation of no. of moles of ClCH2COOH

We have relation, Molarity = No. of moles of solute / Volume of solution in L

no. of moles of ClCH2COOH = Molarity $\times$ Volume of solution in L

no. of moles of ClCH2COOH = 1.40 mol / L $\times$ 0.125 L = 0.175 mol

Step 3 : Calculation of no. of moles of ClCH2COOK

We have, No. of moles of ClCH2COOK = 1.20 $\times$ No. of moles of ClCH2COOH

No. of moles of ClCH2COOK = 1.20 $\times$ 0.175 mol = 0.210 mol

Step 4 : Calculation of mass of ClCH2COOK

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

Mass = No. of moles $\times$ Molar mass

Molar mass of ClCH2COOK = 35.45 + ( 2 $\times$ 12.01 ) + ( 2 $\times$ 1.0079 ) + ( 2 $\times$ 16.00 ) + 39.10 = 132.58 g / mol

mass of ClCH2COOK = 0.210 mol $\times$ 132.58 g / mol = 27.84 g

ANSWER : Mass of ClCH2COOK = 28 g