A 35.00-mL sample contains 1.30 g of KHC_8H_4O_4, known as KHP. This sample is u

Question

A 35.00-mL sample contains 1.30 g of KHC_8H_4O_4, known as KHP. This sample is used to standardize an NaOH solution. At the equivalence point. 50.00 mL of NaOH have been added. K_a(HC_8H_4O_4^-) = 3.9 times 10^-6 Indicators that change color within the pH range: bromothymol blue = 6.0 - 7.7 phenolphthalein = 8.3 - 9.8 alizarin yellow R = 10.0 - 12.0 What was the concentration of the NaOH? Concentration = M What is the pH at the equivalence point? pH = Which indicator, bromothymol blue, phenolphthalein. or alizarin yellow R. should be used in the titration?

Explanation / Answer

a)

mmol of KHP = mass/MW = 1.3/204.22 = 0.006365 mol

so

Ratio is

1:1

1 mol of KHP = 1 mol of base

0.006365 mol of base

[NAOH] = mml/L = 0.006365 / (50*10^-3) = 0.1273 M

b)

find pH in equivalence point

so

Vtotal = 35+50 = 85 mL

mol of KHP --> 0.006365

mmol of KHP forming conjguate base --> 0.006365 mol = 0.006365 *10^3 mmol =6.35 mmol

so

[KHP conjugate] = mmol/mL = 6.35/85 = 0.074705 M

so

KHP hydrolyses as follows:

KP- + H2O <-> KHP + OH- Kb

Kb = Kw/Ka = (10^-14)/(3.9*10^-6) = 2.5641*10^-9

now

Kb = [KHP][OH-]/[KP-]

2.5641*10^-9 = x*x/(0.074705 -x)

x = 1.376*10^-5

[OH-] = 1.376*10^-5

pOH = -log(1.376*10^-5) = 4.861

pH = 14-pOH = 14-4.861 =9.139

c)

indicator --> for a pH range of 8-9, phenolphtalein is a good indicator

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