Suppose 500 mL of 0.200 M chloroacetic acid , which has a p K a of 2.854, soluti

Question

Suppose 500 mL of 0.200 M chloroacetic acid , which has a pKa of 2.854, solution is titrated with 75.0 mL of a 1.00 M NaOH. What is the final pH? The error interval is +/- 0.2 pH units. Answer is 3.33. Show your work.

Suppose you need to create 725 mL of a pH 3.50 solution in water. You have a 0.0250 molar concentrated solution of HCl. How much of this concentrated acid solution will need to be added to a 725 mL volumetric flask and then diluted with water to obtain a final volume of 725 mL at the desired pH? Units are mL.

Thank you

Explanation / Answer

A)

V1 = 500 ml

M1 = 0.2 Acid

pKA = 2.854

V2 = 75

M2 = 1.0 NaOH

Reaction

HA + NaOH ----> H2O + Na+ and A-

From here

Calculate moles reacted

N1 = M1*V1 = 0.2*0.5 = 0.1 mol of acid

N2 = M2*V2 = 1*0.075 = 0.075 mol of base

0.100 -0.075 = 0.025 mol of acid left

This will be:

NaA (aq) ---> Na+ and A-

HA + <---> H+ and A-

There is a common ion effect

this will form a buffer

we decribe buffers with henderson haselbach equation

pH = pKA + log(A-/HA)

Recalculate concentrations of A- and HA (new moles and volumes)

[A-] = 0.075 / (0.500+0.075) = 0.13 M

[HA] left = 0.025 / (0.5+ 0.075) =0.043 M

pH = pKA + log(A-/HA) = 2.854 + log(0.13/0.043) = 3.33

Which is what you got XD

NOTE: please consider posting question number 2 in another set of questions, we are not allowed to answer multiple questions.

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