chemistry

The Ka of hypochlorous acid (HClO) is 3.0 × 10-8 at 25°C. Calculate the pH of a 0.0385-M hypochlorous acid solution?

i got help with this problem, but i don't get where one of the numbers came from (3.4 x 10^-5)??? i tried makeing 3.0 x 10^-8 the x but i still don't get that number.

start
0.0385
change
-x. . . . .. . +x. . . .+x
at equilibrium
0.0385-x. .. x..... . .x

Ka = 3.0 x 10^-8 = [H+][ClO-]/ [HClO]= (x)(x)/ 0.0385-x

x = [H+]= 3.4 x 10^-5 M
pH = - log 3.4 x 10^-5=4.5

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1. Hiya,
From what you said it seems that you understand everything up till "x = [H+]= 3.4 x 10^-5 M"
So we know that 3X10^-8 = x^2/(0.0385-x)
Bring (0.0385-x) over to the left hand side,
Hence 3X10^-8(0.0385-x) = x^2
Then we expand the left hand side,
1.155X10^-9 - 3X10^-8x = x^2
From there we bring all terms to the right hand side to give
x^2 + 3X10^-8x - 1.155X10^-9 = 0
Then we use our quadratic formula to solve for x --> x = [-b +/- sqrt(b^2-4ac)]/2a and reject the negative answer since x = concentration of H+ which cannot be negative. Then we get 3.4X10^-5 M.
Alternatively, if the math is a little too tedious, we can approximate x to be near zero since HClO is a weak acid which only dissociates very little in aqueous solution. Then we get 3X10^-8 = x^2/0.0385 which should be slightly simpler to solve.
Hope I helped!

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2. where is 1.155X10^-9 from?

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3. That comes from multiplying 3X10^-8 and 0.0385 when we expand 3X10^-8(0.0385-x).

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