Question

a) What is the pH of a buffer made by adding 2.91*10^(-2) NaHCO3 and 4.30*10^(-3) M Na2CO3 t owater in a closed system?

b) What is the pH of the buffer after addition of 10^(-3) M H2SO4? The system is still closed.

c) If the bfufer in part a is opened to an atmosphere with Pco2-10^(-3.46) atm, what is the resulting pH? Does Ctco3 increase or decrease, by how much?

Answer 1

Using Henderson-hasselbach,

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

or,pH=pka+log [Na2CO3]/[NaHCO3]

or, pka for CO3^2-/HCO3-=6.35

pH=6.35+log (4.30*10^(-3)/(2.91*10^(-2))=6.35 +(-0.83)=5.52

pH=5.52

b)After addition of 10^(-3) M H2SO4, the concentration of Na2CO3 will be reduced due to neutralization by the acid

2Na2CO3+H2SO4 -----> 2NaHCO3+Na2SO4

Ratio of Concentration of Na2CO3 by H2SO4 reacting =2:1

So change in concentration of Na2CO3=2*[H2SO4]=2* 10^(-3) M

New concentration of Na2CO3=4.30*10^(-3) M - 2* 10^(-3) M=2.30*10^-3 M

concentration of NaHCO3 increase by the same amount as the ratio of Na2CO3 by NaHCO3 is 2:2=1:1

New concentration of NaHCO3= 2.91*10^(-2)+ 2.0*10^(-3)=3.11*10^-2M

pH=6.35+log 2.30*10^-3 M/3.11*10^-2M=6.35-1.131=5.22

pH=5.22

c)pCO2=10^(-3.46) atm,partial pressure

So, -log pCO2

H2O(l) +CO2(g) <--->H2CO3, pka=1.46 =-log ka so ka=10^-1.46=0.0347

ka=0.0347=[H2CO3]eq/pCO2

[H2CO3]eq=0.0347*pCO=1.097*10^-5M

Thus ,[CO32-]=4.30*10^(-3) M+1.097*10^-5M =increase in concentration as H2CO3 is formed by CO2+H2O reaction adds to carbonate concentration

negligible change so ,pH =5.52(prt a)