# pH problem

#### inaccensa

10+ Year Member
7+ Year Member
If the pKa1 value for H2CO3 is 6.4, what is the pH of
a 103 M solution of this acid?
A . 3.0
B . 4.7 ---correct ans
C . 6.3
D . 9.3

Ka will be 10^-6.4 and thus Ka = (HCO3-)(H+)/ (H2CO3)
10^-6.4 = x2/ 10^-3
x2 = 10^9.4
x approximately equal to 3, infact a little less than 3.

Can anyone please solve this problem. I don't understand the Kaplan explanation. They said that the ka = 4*10^-7

#### RogueUnicorn

##### rawr.
7+ Year Member
If the pKa1 value for H2CO3 is 6.4, what is the pH of
a 103 M solution of this acid?
A . 3.0
B . 4.7 ---correct ans
C . 6.3
D . 9.3

Ka will be 10^-6.4 and thus Ka = (HCO3-)(H+)/ (H2CO3)
10^-6.4 = x2/ 10^-3
x2 = 10^9.4
x approximately equal to 3, infact a little less than 3.

Can anyone please solve this problem. I don't understand the Kaplan explanation. They said that the ka = 4*10^-7
you did it correctly... what's the problem?

x2 = 10^-9.4 ~~ x2 = 10^-10
x = 10^-5
pH=5 ~~4.7

#### dragon529

##### MS-III
10+ Year Member
a shortcut for finding pH of weak acids

pH = 1/2 * pKa - 1/2 * log[H+]

Plugging those values in will give you 4.7

#### crazybob

h2co3 + h2o <-> hco3- + h3o+
--------------------------------
10^-3...-----........0...........0
...-x......-----........+x........+x
(10^-3)-x----.......x..........x

Ka1 = x^2 / (10^-3 - x)
and solve for x = [H3O+]
pH = -log x

Last edited:
OP

#### inaccensa

10+ Year Member
7+ Year Member
you did it correctly... what's the problem?

x2 = 10^-9.4 ~~ x2 = 10^-10
x = 10^-5
pH=5 ~~4.7

Thats what I did, I rounded that 10^-9.4 to 10^-10 and then u get 4.7, but technically isn't that wrong, since it should have been 9.5 to round that off to 10. Is there a way that you can actually solve it without rounding that off?

OP

#### inaccensa

10+ Year Member
7+ Year Member
a shortcut for finding pH of weak acids

pH = 1/2 * pKa - 1/2 * log[H+]

Plugging those values in will give you 4.7

did you mean pH = 1/2 * pKa + 1/2 * log[H+]

#### RogueUnicorn

##### rawr.
7+ Year Member
Thats what I did, I rounded that 10^-9.4 to 10^-10 and then u get 4.7, but technically isn't that wrong, since it should have been 9.5 to round that off to 10. Is there a way that you can actually solve it without rounding that off?
no it's not wrong. the MCAT doesn't expect you to mentally calculated 10^-9.4... even if you round it to 10^-9, then x=10^-4.5, so pH ~~ 4.5; again, close enough.

OP

#### inaccensa

10+ Year Member
7+ Year Member
no it's not wrong. the MCAT doesn't expect you to mentally calculated 10^-9.4... even if you round it to 10^-9, then x=10^-4.5, so pH ~~ 4.5; again, close enough.

ok, but that explanation from kaplan was a mess. how do you end up Ka = 4*10^-7?

#### RogueUnicorn

##### rawr.
7+ Year Member
ok, but that explanation from kaplan was a mess. how do you end up Ka = 4*10^-7?
unsure.. 10^-9.4 does equal to about 4*10^-10

OP

10+ Year Member
7+ Year Member

#### kentavr

10+ Year Member
no it's not wrong. the MCAT doesn't expect you to mentally calculated 10^-9.4... even if you round it to 10^-9, then x=10^-4.5, so pH ~~ 4.5; again, close enough.
You actually do not need rounding here since
x^2 = 10^-9.4
x = 10^(-9.4/2) = 10^-4.7
pH = 4.7

#### RogueUnicorn

##### rawr.
7+ Year Member
You actually do not need rounding here since
x^2 = 10^-9.4
x = 10^(-9.4/2) = 10^-4.7
pH = 4.7

haha true that

#### HopefulOncoDoc

10+ Year Member
5+ Year Member
did you mean pH = 1/2 * pKa + 1/2 * log[H+]
No, he's right. For weak acids: pH = 1/2 pKa - 1/2 log [HA]

pH = 1/2(6.4) - 1/2( log [10^-3]) = 3.2 - 1/2 (-3 - log1) = 3.2 - 1/2 (-3)
= 3.2 + 1.5 = 4.7