Question

Uctober___ ,2019 Name 1) Consider the equilibrium HBB(aq) < > BB(aq) + H(aq) What will be ratio [HBB]/[BB'] at pH=7.10 K(a) = 2.0 10" 2) Calculate pH of a buffer solution prepared by mixing 1000.0mL of 0.100 M NH; with 14.80 g of solid NH4Cl (molar mass-53.50g/mol). KB(NH3)=1.8 10-3

Answer 1

Consider reaction ,HBB (aq) $\rightleftharpoons$ BB ^- (aq) + H ⁺ (aq)

For above reaction, Ka = [BB ^- ] [H ⁺] / [HBB] = 2.0 $\times$ 10 ^-07

We have, pH = -log [H ⁺]

$\therefore$ [H ⁺] = 10 ^-pH = 10 ^-7.10 = 7.943 $\times$ 10 ^-08 M

Putting [H ⁺] = 7.943 $\times$ 10 ^-08 M in above equation we get

2.0 $\times$ 10 ^-07 = [BB ^- ] 7.943 $\times$ 10 ^-08 / [HBB]

[BB ^- ]  / [HBB] = 2.0 $\times$ 10 ^-07 / 7.943 $\times$ 10 ^-08 = 2.52

$\therefore$ [HBB] / [BB ^- ] = 1 /2.52 = 0.397

ANSWER : [HBB] / [BB ^- ] = 0.397

We have Henderson's equation for basic buffer solution pOH = pKb + log [Salt] / [Base]

$\therefore$ pOH = pKb + log [NH₄Cl] / [NH₃]

We have, pKb = - log Kb = - log (1.8 $\times$ 10 ^-05 ) = 4.74

[NH₃] = 0.1000 M

[NH₄Cl] = No. of moles of NH₄Cl / Volume of solution in L

[NH₄Cl] = ( Mass of NH₄Cl / Molar mass of NH₄Cl) / Volume of solution in L

$\therefore$ [NH₄Cl] = ( 14.80 g / 53.50 g/mol ) / 1.000 L

[NH₄Cl] = 0.2766 mol / 1.000 L = 0.2766 M

$\therefore$ pOH = 4.74 + log 0.2766 / 0.1

pOH = 4.74 + 0.4418

pOH = 5.18

We know that, pH + pOH = 14

Therefore, pH = 14- pOH = 14 - 5.18 = 8.82

ANSWER : pH of buffer solution = 8.82