10. Calculate the pH before and after adding 0.05 mol of NaOH to (1) 1.00 L H>0,

Question

10. Calculate the pH before and after adding 0.05 mol of NaOH to (1) 1.00 L H>0, (2) 1.00 L buffer made from 0.45 M HF (Ka = 7.2x 10") and 0.50 M NaF, and (3) 1.00 L buffer made from 0.90 M HF (Ka = 7.2x 104) and 1.00 M NaF pH 1.00-L H20 1.00-L Buffer 1.00-L Buffer (0.45 M HF/0.50 M NaF) (0.90 MHF/1,00 M NaF) Before adding 0.05 mol OH After adding 0.05 mol OH pH change 11. You are going to add solid CH,COONa to 1000.-mL 0.60 M CH,COOH (K, 1.8x 109) to make a pH nf the salution does not change when adding solid CH COONa,

Explanation / Answer

a)

before additio of base

i) water = pH = 7

ii) diluted buffer

pH = pKa + log(NaF/HF)

pH = 3.15 + log(0.5/0.45) = 3.20

ii) concentrated buffer

pH = pKa + log(NaF/HF)

pH = 3.15 + log(1/0.90) = 3.20

b)

after addition of OH-

i) water

[OH-] = 0.05/1 = 0.05

pH = 14 + log(OH) = 14+ log(0.05) = 12.70

ii) diluted buffer

pH = pKa + log(NaF/HF)

pH = 3.15 + log((0.5+0.05)/(0.45-0.05)) = 3.29

ii) concentrated buffer

pH = pKa + log(NaF/HF)

pH = 3.15 + log((1+0.05)/(0.90-0.05)) = 3.24

c)

pH changes

i) water = pH = 7

pH = 12.70-7= 5.70

ii) diluted buffer

dpH = 3.29-3.20 = 0.09

iii) concentrated

dpH = 3.24-3.20 = 0.04

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