Ka=[H][COOH]/([HCOOH]
ka= x^2/(.5-x)
solve this for x
x^2+.00018x-.5*.00018=0
x=(-.00018+-sqrt(.00018^2+.00036))/2
x=-.00009+- .00948= .00939 check all that math.
pH=-log (.00939)=2.03
check the math.
Calculate the pH of a 0.5 M Formic Acid (HCOOH) soln. ka= 0.00018
3 answers
The equilibrium is given by
HCOOH -> H+ + HCOO-
so Ka=[H+][HCOO-]/[HCOOH]
if we start with 0.5M HCOOH then at equilibrium there is [H+] = x and [HCOO-]=x and HCOOH=0.5-x
so Ka = (x)(x)/(0.5-x)=0.00018
this gives a quadratic to solve, however, we only want the answer to 1 sig fig (.5M is 1 sig fig) so we can say that 0.5-x is approximately equal to 0.5, because x will be small.
thus x^2/0.5=0.00018
hence you can find x
pH is then -log(x)
which is pH=2, but check the maths.
HCOOH -> H+ + HCOO-
so Ka=[H+][HCOO-]/[HCOOH]
if we start with 0.5M HCOOH then at equilibrium there is [H+] = x and [HCOO-]=x and HCOOH=0.5-x
so Ka = (x)(x)/(0.5-x)=0.00018
this gives a quadratic to solve, however, we only want the answer to 1 sig fig (.5M is 1 sig fig) so we can say that 0.5-x is approximately equal to 0.5, because x will be small.
thus x^2/0.5=0.00018
hence you can find x
pH is then -log(x)
which is pH=2, but check the maths.
lasaga
1 answer