Question

Calculate the pH of the solution that results from mixing 61.0mL of 0.052 MHCN(aq) with 39.0 mL of 0.025 M NaCN(aq).The ?a value for HCN is 4.9×10−10

pH=

Calculate the pH of the solution that results from mixing 30.0 mL of 0.027 M HCN(aq) with 70.0 mL of 0.064 M NaCN(aq).

pH=

Calculate the pH of the solution that results from mixing 42.0 mL of 0.103 M HCN(aq) with 42.0 mL of 0.103 M NaCN(aq).

pH=

Answer 1

1)

Concentration after mixing = mol of component / (total volume)

M(NaCN) after mixing = M(NaCN)*V(NaCN)/(total volume)

M(NaCN) after mixing = 0.025 M*39.0 mL/(39.0+61.0)mL

M(NaCN) after mixing = 9.75*10^-3 M

Concentration after mixing = mol of component / (total volume)

M(HCN) after mixing = M(HCN)*V(HCN)/(total volume)

M(HCN) after mixing = 0.052 M*61.0 mL/(61.0+39.0)mL

M(HCN) after mixing = 3.172*10^-2 M

Ka = 4.9*10^-10

pKa = - log (Ka)

= - log(4.9*10^-10)

= 9.31

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.31+ log {9.75*10^-3/3.172*10^-2}

= 8.797

Answer: 8.80

2)

Concentration after mixing = mol of component / (total volume)

M(NaCN) after mixing = M(NaCN)*V(NaCN)/(total volume)

M(NaCN) after mixing = 0.064 M*70.0 mL/(70.0+30.0)mL

M(NaCN) after mixing = 4.48*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(HCN) after mixing = M(HCN)*V(HCN)/(total volume)

M(HCN) after mixing = 0.027 M*30.0 mL/(30.0+70.0)mL

M(HCN) after mixing = 8.1*10^-3 M

Ka = 4.9*10^-10

pKa = - log (Ka)

= - log(4.9*10^-10)

= 9.31

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.31+ log {4.48*10^-2/8.1*10^-3}

= 10.05

Answer: 10.1

3)

Concentration after mixing = mol of component / (total volume)

M(NaCN) after mixing = M(NaCN)*V(NaCN)/(total volume)

M(NaCN) after mixing = 0.103 M*42.0 mL/(42.0+42.0)mL

M(NaCN) after mixing = 5.15*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(HCN) after mixing = M(HCN)*V(HCN)/(total volume)

M(HCN) after mixing = 0.103 M*42.0 mL/(42.0+42.0)mL

M(HCN) after mixing = 5.15*10^-2 M

Ka = 4.9*10^-10

pKa = - log (Ka)

= - log(4.9*10^-10)

= 9.31

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.31+ log {5.15*10^-2/5.15*10^-2}

= 9.31

Answer: 9.31