Solution 3 contains 24.0mL of 1.0M NaOAc + 10.0mL of 1.0 M HOAc Solution 4 conta

Question

Solution 3 contains 24.0mL of 1.0M NaOAc + 10.0mL of 1.0 M HOAc
Solution 4 contains 10.0mL of the above mixture + 40.0mL of water. What is solution 4 Ka Measured pH for solution for is 5.00 I tried to comment or edit my last attempt and couldn't figure it out so I posted it again. Solution 3 contains 24.0mL of 1.0M NaOAc + 10.0mL of 1.0 M HOAc
Solution 4 contains 10.0mL of the above mixture + 40.0mL of water. What is solution 4 Ka Measured pH for solution for is 5.00 I tried to comment or edit my last attempt and couldn't figure it out so I posted it again.
Solution 4 contains 10.0mL of the above mixture + 40.0mL of water. What is solution 4 Ka Measured pH for solution for is 5.00 I tried to comment or edit my last attempt and couldn't figure it out so I posted it again.

Explanation / Answer

Solution:- HOAc is a weak acid and NaOAc is it's salt so it will work as a buffer solution.

initial mmol of HOAc = 10.0 x 1.0 = 10.0 mmol

initial mmol of NaOAc or OAc- = 24.0 x 1.0 = 24.0 mmol

total volume = 24.0 ml + 10.0 ml = 34.0 ml

10.0 ml of this original solution are taken.

so, the mmol of HOAc present in 10.0 ml of the buffer = 10.0 ml x (10.0 mmol/34.0 ml) = 2.94 mmol

mmol of OAc- present in 10.0 ml of the buffer = 10.0 ml x (24.0 mmol/34.0 ml) = 7.06 mmol

total volume of solution after adding 40.0 ml of water to the 10.0 ml of the buffer = 40.0 ml + 10.0 ml = 50.0 ml

[HOAc] = 2.94 mmol/50.0 ml = 0.0588 M

[OAc-] = 7.06 mmol/50.0 ml = 0.1412 M

pH of the the solution is given as 5.00

From Handerson equation, pH = pKa + log (A-/HA)

5.00 = pKa + log(0.1412/0.0588)

5.00 = pKa + 0.38

pKa = 5.00 - 0.38

pKa = 4.62

Ka = 10-pKa

Ka = 10-4.62

Ka = 2.40 x 10-5

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