Question

calculate the ph of 1.0 l of a buffer that is .01 m of HNO2 and .15 m NaO2. what is the ph of the same buffer and after the additon of 1.0 l of 12m hcl (pka of HNO2 is 3.4)

Answer 1

its an acidic buffer as HNO2 is a weak acid and NaNO2 is a salt of weak acid HNO2 and strong base NaOH

pH = pKa + log [salt] / [acid]

[salt] = [NaNO2] = 0.15 M
[acid] = [HNO2] = 0.1M

pH = 3.4 + log 0.17/0.1

pH = 3.4 + log 1.7 = 3.4 + 0.23 = 3.63

now when you add HCl following reaction takes place ...

NaNO2 + HCl -----> HNO2 + NaCl

so HCl converts NaNO2 into HNO2 ....in other words it decreases concentration of NaNO2 and increases concentration of HNO2 ...

initially no. of moles of HNO2 = molaity X volume of solution in litres = 0.1 X 1 = 0.1
similarly no. of moles of NaNO2 = 0.15 X 1 = 0.15

now no. of moles of HCl added = 12 X 1/1000 = 0.012

since HCl and NaNO2 are reacting in 1:1 ratio ....

so 0.012 moles of HCl will convert 0.012 moles of NaNO2 into 0.012 moles of HNO2...

so final no. of moles of HNO2 = 0.1 + 0.012 = 0.112
final no. of moles of NaNO2 = 0.15 - 0.012 = 0.138

total volume = 1 + 1/1000 = 1.001 L

[salt] = [NaNO2] = 0.138 / 1.001 = 0.138 M
[acid] = [HNO2] = 0.112/1.001 = 0.1118 M

pH = 3.4 + log 0.138/0.112

pH = 3.4 + log 1.232 = 3.4 + 0.091 = 3.4906