Question

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 mL of 0.25 M HF with 225.0 mL of 0.30 M NaF b. 175.0 mL of 0.10 M C H NH, with 275.0 mL of 0.20 M CH,NH4Cl 17. Calculate the ratio of NaF to HF required to create a buffer with pH = 4.00. IT YILC concentration

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