Question

A 200.00mL sample of 1.000M nitric acid is added to a carbonate buffer solution prepared by mixing 1.00mol Na2CO3 and 1.000 mol NAHCO3 and adding water to make a 1.000L solution. What is the resulting pH? (pKa of HCO3- is 10.32)

Answer 1

original pH :

pKa of HCO₃^- = 10.32

[Na2CO3] = moles / volume = 1.000 mol/ 1.000 L = 1.0 M

[NaHCO3] = moles / volume = 1.000 mol/ 1.000 L = 1.0 M

According to Henderson-Hasselbalch equation,

pH = pKa + log ([Na2CO3] / [NaHCO3] )

= 10.32+ log (1.0 M/1.0 M)

= 10.32 + 0

= 10.32

pH = 10.32

pH after addition of  200.00mL sample of 1.000M nitric acid

Moles of HNO3 = molarity x volume in Litres = 1.000 M x 0.200 L = 0.2 mol

Moles of Na2CO3 = 1.000 mol

Moles of NaHCO3 = 1.000 mol

Na2CO3 + HNO3 ---------------> NaHCO3 + NaNO3

1.000 mol 0.2 mol 0 mol

----------------------------------------------------------------------------

1.0-0.2 0 0.2 mol

= 0.8 mol

Therefore,

[Na2CO3] = 0.8 mol

[NaHCO3] = Initial moles + 0.2 mol = 1.0 mol + 0.2 mol = 1.2 mol

pKa of HCO₃^- = 10.32

pH = pKa + log ([Na2CO3] /  [NaHCO3] )

= 10.32+ log (0.8 /1.2)

= 10.14

pH = 10.14

Therefore, pH after addition of  200.00 mL sample of 1.000M nitric acid = 10.14