a) What is the pH of a buffer
a) What is the pH of a buffer made by adding 2.91*10^(-2) NaHCO3and 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) MH2SO4? The system is still closed.
c) If the bfufer in part a is opened to an atmosphere withPco2-10^(-3.46) atm, what is the resulting pH? Does Ctco3 increaseor decrease, by how much?
Answer:
Using Henderson-hasselbach,
pH=pKa+log [base]/[acid]
or,pH=pka+log [Na2CO3]/[NaHCO3]
or, pka for CO32-/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 Na2CO3will 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 ratioof 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 soka=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 inconcentration as H2CO3 is formed by CO2+H2O reaction adds tocarbonate concentration
negligible change so ,pH =5.52(prt a)
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