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April 18, 2015

April 18, 2015

Posted by **Anonymous** on Thursday, November 4, 2010 at 10:32am.

- College Chemistry -
**bobpursley**, Thursday, November 4, 2010 at 10:52amKa=[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.

- College Chemistry -
**Dr Russ**, Thursday, November 4, 2010 at 10:52amThe 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.

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