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lizzyb
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Question
A buffer solution is prepared by mixing 600.0 ml of 0.600 M HClO and 400.0 ml of 1.00 M NaClO. Calculate the pH of the solution after 40.0 ml of 3.00 M HCl is added to the buffer. For HClO, Ka = 3.0 X 10^-8.
Work so Far
does this look right to you?
A buffer solution is prepared by mixing 600.0 ml of 0.600 M HClO and 400.0 ml of 1.00 M NaClO. Calculate the pH of the solution after 40.0 ml of 3.00 M HCl is added to the buffer. For HClO, Ka = 3.0 X 10^-8.
Work so Far
Code:
A buffer solution os prepared by mixing 600.0 ml of 0.600 M HClO and
400.0 ml of 1.00 M NaClO. Calculate the pH of the solution after
40.0 ml of 3.00 M HCl is added to the buffer. For HClO,
Ka = 3.0 X 10^-8.
Work so Far:
Initial Concentration of the Buffer Solution: Now NaClO dissolves to completion so we have an initial
concentration of ClO:
1 M * .4 L
[ClO]_initial: ------------- = .4 M
(.6 + .4)L .6 M * .6 L
[HClO]_initial: ------------- = .36 M
(.6 + .4)L HClO <---> H+ + ClO-
---- -- ----
.36 0 .4 initial
-x x x change
.36-x x .4+x equilibrium x(.4+x)
so Ka = --------- ==> x = 1.206 X 10^-5
(.36-x)
which is [H+] (its concentration) and for ClO-,
its new concentration is .400012 in 1 Liter,
thus there are .4 moles of ClO now.
After HCl is added:
Moles of HCl added: 3.00M * .04 L = .12 moles
These .12 moles will react with the .4 moles of ClO
and will be used up so there will be no change in
the number of moles of H+.
There is, however, a change in concentration. For if
we have a concentration of 1.206 X 10^-5 M for H+,
then our new concentration is:
-5
1.206 X 10 * 1 L -5
[H+] = --------------------- = 1.15954 X 10
(1 + .04) L
pH = -log([H+]) = 4.936
does this look right to you?
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