Calculate the pH of a 1,638 mL solution containing 0.45 moles of carbonic acid
4 answers
Is it 3.46?
Post your work so I can check it.
I used an online calculator. Our teacher didn't explain this well on how to work pH
H2CO3 is a weak diprotic acid. Since k2 is so small (about 10^-11) I will go only with k1 = 4.0E-7
Molarity of H2CO3 in the solution is mols/L = 0.45/1.638 = 0.275 M
.....................H2CO3 ==> H^+ + HCO3^-
I.....................0.275............0.............0
C......................-x................x.............x
E................0.275-x..............x............x
k1 = 4.0E-7 = (H^+)(HCO3^-)/(H2CO3) = (x)(x)/0.275-x)
Solve for x. First I make the assumption that 0.275-x = 0.275 since x is small.
x^2 = 0.3275*4.0E-7 and x = 3.32E-4 = (H^+)
Then pH = -log(3.32E-4) = 3.47
So your on-line calculator was correct; however, you can't depend on those. You must learn how to do these yourself. My solution above will show you how.
Molarity of H2CO3 in the solution is mols/L = 0.45/1.638 = 0.275 M
.....................H2CO3 ==> H^+ + HCO3^-
I.....................0.275............0.............0
C......................-x................x.............x
E................0.275-x..............x............x
k1 = 4.0E-7 = (H^+)(HCO3^-)/(H2CO3) = (x)(x)/0.275-x)
Solve for x. First I make the assumption that 0.275-x = 0.275 since x is small.
x^2 = 0.3275*4.0E-7 and x = 3.32E-4 = (H^+)
Then pH = -log(3.32E-4) = 3.47
So your on-line calculator was correct; however, you can't depend on those. You must learn how to do these yourself. My solution above will show you how.
3 answers