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29. E 2.- b. 125.0 mL of 0.10 M NH; WILT JUUM 16. Calculate the pH of the solution that results from each mixture. a. 150.0 m
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Answer #1

16)

a) After mixing the solutions

Concentration of HF = 0.25M/(375ml/150ml) = 0.10M

Concentration of F-= 0.30M / ( 375ml/225ml) = 0.18M

pKa of HF = 3.17

Applying Henderson - Hasselbalch eqution

pH = pKa + log([A-] /[HA])

pH = 3.17 + log( 0.18M/0.10M)

pH = 3.17 + 0.26

pH = 3.43

b) After mixing the solutions

concentration of C2H5NH2 = 0.10M/( 450ml / 175ml) = 0.03889M

concentration of C2H5NH3+ = 0.20M/(450ml/275.0ml ) = 0.1222M

pKa of C2H5NH3+ = 10.8

Applying Henderson - Hasselbalch equation

pH = pKa + log([A-]/[HA])

pH = 10.8+ log( 0.03889M/ 0.1222M)

pH = 10.8 - 0.50

pH = 10.30

17)

Applying Henderson- Hasselbalch equation

pH =pKa + log([A-]/[HA])

pH = pKa + log([F-]/[HF])

4.00 = 3.17 + log([F-]/[HF])

log([F-]/[HF]) = 0.83

[F-]/[HF] = 6.76

[NaF]/[HF] = 6.76

Therfore ,

ratio of NaF to HF = 6.76

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