Question

I'm stuck at the end of this problem!!
"Calculate the pH of a soln prepared by mixing 300 mL of 0.25 M sodium hydrogen ascorbate and 150 mL of 0.2 HCl plus water to a final volume of 1 L. The Ka of ascorbic acid is 5x10^(-5). "

Answer 1

Here is what I've done so far:

pKa= -logKa
pKa= -log (5E10-5)
pKa = 4.3

Answer 2

pH = pKa + log [A-]/[HA]
pH=4.3 = log [75/30]

pH=x
10^(-pH)= [H+]i in 450 mL

C1V1=C2V2
[H+]i [450mL] = [H+]f [1000mL] <--(i = initial and f = final)

Answer 3

oh and I changed mL to mmol...that's where I got the 75 from

Answer 4

So I know that [H+]f = ([H+]i [450])/[1000) <-(i = initial)
AND i know that pHf = - log [H+]f

Answer 5

but I get stuck after the c1v1 part

Answer 6

Is there a reaction between the substances that are introduced in water ?

If so, all problems of this type must start by writing down the equation of the reaction that is taking place.

Answer 7

There is no reaction between the substances. The work I did is right, I'm just not sure how to proceed.

Answer 8

Sorry to disagree, but HCl is a strong acid and hydrogenascorbate a weak base.
So there *is* a reaction taking place.

I do not understand your pH = 4.3 + log [75/30]
What does the 75 represent?

Answer 9

I converted into millimoles. This is the problem I was given by my teacher, I am not to do anything but the math portion of the problem. And I apologize, I was not to do c1v1, I was to do an ICE chart

Answer 10

I do not know what your ICE chart looks like, but it will be useless if you start on the false assumption that no reaction is taking place when you mix your reactants.

There IS a total reaction taking place. Write it down, balance it, find amounts in the solution, and only then, with the new amounts, can you work out the value of the pH (no further ICE chart needed in this example).