Question

5. (28 pts) Calculate the pH after the addition of 20.00, 35.00, and 50.00 mL of 0.400 M HCI to a solution that contains 100.00 mL of 0.100 M Na,SO, H,SO,(aq) + H 20(1) HSOİ (aq) + H 30' (aq) K1.23 x 10 al K." 6.6 × 10-8

Answer 1

Hi

Let us taken it in three cases.

Case 1 – 100 ml H2SO3 +20 ml HCl

We should remember that the H2SO3 is a weak acid and HCl is a strong acid.

Let us see the two reactions –

H2SO3+H2O = HSO3- + H3O+     ka= 1.23 *10 ^-2

HSO3- + H2O= SO3 ^2- + H3o+        ka2= 6.6*10 ^-8

Moles of H2SO3 = 0.1*0.1=0.01 moles

Part a -

Moles of HCl= 0.4*20= 0.008 moles

Volume =120 ml

HCl=H+ + OH-

This will dissociate 100%

So moles of H3O+ from HCL = moles of H+ =0.008 moles

H2SO3+H2O = HSO3- + H3O+    

Let the moles reacted be x .

H2SO3 = 0.01 moles –x moles

Moles of HSO3- = x

Moles of H3O+    =0.008+x

H2O is abundant.

Ka= [HSO3-][ H3O+ ]/ H2SO3=(0.008+x)* x/( 0.01-x) =1.23 *10 ^-2

X2+0.008X -0.000123+0.0123X=0

X2+0.020X -0.000123=0

X=0.0049 moles

So H30+= 0.008+0.0049 = 0.013 moles

As the ka2 value is very less as compared to ka1,so the moles coming out will be negligible.

Volume will be 120 ml

Concentration = moles/volume = 0.013 moles/0.12 = 0.108

pH= -log[H+]

=1

Part b -

Moles of HCl= 0.4*0.035= 0.014 moles

Volume =135 ml

HCl=H+ + OH-

This will dissociate 100%

So moles of H3O+ from HCL = moles of H+ =0.014 moles

H2SO3+H2O = HSO3- + H3O+    

Let the moles reacted be x .

H2SO3 = 0.01 moles –x moles

Moles of HSO3- = x

Moles of H3O+    =0.014+x

H2O is abundant.

Ka= [HSO3-][ H3O+ ]/ H2SO3=(0.014+x)* x/( 0.01-x) =1.23 *10 ^-2

X2+0.014X -0.000123+0.0123X=0

X2+0.026X -0.000123=0

X=0.0040 moles

So H30+= 0.014+0.0040 = 0.018 moles

As the ka2 value is very less as compared to ka1,so the moles coming out will be negligible.

Volume will be 135 ml

Concentration = moles/volume = 0.018 moles/0.135 = 0.133

pH= -log[H+]

=0.87

Part c

Moles of HCl= 0.4*0.05= 0.02 moles

Volume =150 ml

HCl=H+ + OH-

This will dissociate 100%

So moles of H3O+ from HCL = moles of H+ =0.02 moles

H2SO3+H2O = HSO3- + H3O+    

Let the moles reacted be x .

H2SO3 = 0.02 moles –x moles

Moles of HSO3- = x

Moles of H3O+    =0.02+x

H2O is abundant.

Ka= [HSO3-][ H3O+ ]/ H2SO3=(0.02+x)* x/( 0.01-x) =1.23 *10 ^-2

X2+0.02X -0.000123+0.0123X=0

X2+0.0332X -0.000123=0

X=0.0033 moles

So H30+= 0.02+0.0033 = 0.023 moles

As the ka2 value is very less as compared to ka1,so the moles coming out will be negligible.

Volume will be 150 ml

Concentration = moles/volume = 0.0173 moles/0.15 = 0.153

pH= -log[H+]

=0.81

Please note that we have taken a approximation that moles of H3O+ coming from the third reaction is negligible so the answer may be slightly different .

Feel free to comment in the comment section in the case of any doubt.

Thank You