Achieve week 2 number 1

[Daniel N]

So for this problem I know what I'm supposed to do. I am stuck on solving for x
so far I have
x^2+ 2.1E-6x - 2.94E-7

to find the pH I solve pH=-log[H+]
for X I got -1.05E-6
when i plug it in i get 5.98 as the pH
I'm not sure what is wrong

[AudreyQian1J]

Hello,

I believe you may have had a calculation error when doing the quadratic formula. When plugging in the numbers based on your formula x^2+ 2.1E-6x - 2.94E-7, we get [-(2.1E-6)+/-(sqrt[(2.1E-6)^2]-4(1)(-2.94E-7))/2(1)], so x=9.039E-4 and x=-9.060E-4. Since there is no such thing as a negative concentration, we ignore the negative x value and use 9.039E-4 as our concentration for H3O+.
pH = -log[9.039E-4] = 3.04

I hope this helps.

[Shannon Lau 14B - 1H]

For this problem, because the given Ka is so small we could assume that X won't really change the equilibrium at the denominator of HA, so Ka= x^2/0.27. And because Ka < 10^-3, we don't have to use the quadratic formula and can try to solve it by multiplying the denominator to the Ka and finding the square root of it.
After that we must check to see if it is less than 5% so we do x/denominator - which is the initial of HA. Multiply that by 100 to get the percentage and it is <5% so we can use the x that we got.
Then find the pH = -log(x) which should = the correct answer.

I hope this helps!

[Daniel N]

Hello,

I believe you may have had a calculation error when doing the quadratic formula. When plugging in the numbers based on your formula x^2+ 2.1E-6x - 2.94E-7, we get [-(2.1E-6)+/-(sqrt[(2.1E-6)^2]-4(1)(-2.94E-7))/2(1)], so x=9.039E-4 and x=-9.060E-4. Since there is no such thing as a negative concentration, we ignore the negative x value and use 9.039E-4 as our concentration for H3O+.
pH = -log[9.039E-4] = 3.04

I hope this helps.

Thank You Audrey! It did help. I also realized whenever I write down the problem out fully I tend to not make silly mistakes.