Asked by jaycab
COMMON ION EFFECT
1.) What is the pH of a solution that is 0.30M in HCOOH and 0.52M in HCOOK.
Ka of HCOOH=1.7x10^-4.
2.) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0M CH3COONa. Ka of CH3COOH=1.8x10^-5.
1.) What is the pH of a solution that is 0.30M in HCOOH and 0.52M in HCOOK.
Ka of HCOOH=1.7x10^-4.
2.) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0M CH3COONa. Ka of CH3COOH=1.8x10^-5.
Answers
Answered by
Dr Russ
1)
I notice that your value for Ka in the question is missing units, were these missing in the original question?
Start from the equation for the equilibrium reaction
HCOOH<->H+ + HCOO-
at start
0.30M
at equilibrium if [H+]=x
0.30-x x x
there is also 0.52M HCOO-
so at equilibrium
0.30-x x x+0.52
Ka=[H+][HCOO-]/[HCOOH]
Ka=(x)(x+0.52)/(0.30-x)=1.7x10^-4
(x)(x+0.52)/(0.30-x)=1.7x10^-4
You can either solve the quadratic or we can say that if x is small with respect 0.52 abd 0.3, we can rewrite the expression as
(x)(0.52)/(0.30)=1.7x10^-4
and find x
pH is then -log (x/mol litre^-1)
I notice that your value for Ka in the question is missing units, were these missing in the original question?
Start from the equation for the equilibrium reaction
HCOOH<->H+ + HCOO-
at start
0.30M
at equilibrium if [H+]=x
0.30-x x x
there is also 0.52M HCOO-
so at equilibrium
0.30-x x x+0.52
Ka=[H+][HCOO-]/[HCOOH]
Ka=(x)(x+0.52)/(0.30-x)=1.7x10^-4
(x)(x+0.52)/(0.30-x)=1.7x10^-4
You can either solve the quadratic or we can say that if x is small with respect 0.52 abd 0.3, we can rewrite the expression as
(x)(0.52)/(0.30)=1.7x10^-4
and find x
pH is then -log (x/mol litre^-1)
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