Show stepwise how you would prepare 500. ml of a pH 2.00 buffer. Also show the b

Question

Show stepwise how you would prepare 500. ml of a pH 2.00 buffer. Also show the buffer capacity by determining the pH change upon adding 10.0 mL 0.100 M HCl. Suppose that 0.483 g of an unknown monoprotic weak acid. HA. is dissolved in 35 mL ot water. Titration of the solution with 0.250 M NaOH required 42.0 mL to reach the equivalence point. What is the molar mass of the acid? After the addition of 21.0 mL of NaOH. the pH of the solution was found to be 3.75. What is the pK_a of the acid? Use the information determined from parts (a) and (b) to calculate the pH alter the addition of 0 mL NaOH. 10 ml. NaOH, and 25 mL NaOH.

Explanation / Answer

1. (i)Preparation of 500ml of pH=2 buffer.
For buffer of pH=2, pKa of acid should be around 2.
sulfurous acid H5IO6 has Ka1=1.23E-2, Ka2=6.6E-8
pKa1 = 1.9, pKa2=7.2
pH=2, [H3O+]=10^-2=0.01

Buffer Equilibria:
[HA]=[H3O+]+[A-]
Ka = [H3O+][A-]/[HA]
[A-]/[HA] = Ka1/[H30+]= 1.23E-2/0.01=1.23

Let x ml of 0.1M sulfurous acid, [HA] added and
volume of 0.1M sodium hydrogen sulfite, (salt of sulfurous acid) [A-] added = (500-x)ml

[A-]/[HA] = (500-x)*0.1/(x*0.1) = 1.23
(500 - x)/x = 1.23
x = 224.2ml; 500-x = 275.8ml
So 224.2 ml of 0.1M sulfurous acid and 275.8ml of 0.1M sodium hydrogen sulfite is required

(ii) Buffer capaciy

Buffer of 224.2 ml of 0.1M sulfurous acid and 275.8ml of 0.1M sodium hydrogen sulfite

10ml of 0.1M HCl added will react with the basic part [A-].
[A-] + [H30+] = [HA]

[HA]   = 0.1*224.2/500 = 0.045 M
[A-]   = 0.1*275.8/500 = 0.055 M
[H3O+] = 10^(-pH) = 10^(-2) = 0.01 M

Initial:
10ml of 0.1M HCl will decrease the amount of [A-] by 10*0.1 = 1 millimol
and increase the amount of [HA] by 1 millimol

[HA] = (0.045*500+1)/(500+10) = 0.046 M
[A-] = (0.055*500-1)/(500+10) = 0.052 M

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