Question

A 1.00 L solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA. What...

A 1.00 L solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA.

What is the pH of this solution?
What is the pH if 0.310 L of 0.138 M KOH is added to the solution be?
What is the pH if 0.620 L of 0.138 M KOH is added to the solution be?
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Given solution contain weak acid HA and its salt NaA. Therefore this solution is a buffer solution. We can use Henderson's equation to calculate pH of buffer solution.

We have Henderson's equation, pH = pKa + log [ Salt ] / [ acid ]

\therefore pH = - log Ka + log [ NaA ] / [ HA ]

\therefore pH = - log 4.66 \times 10 -06 + log 0.758 / 0.352

pH = 5.332 + log 0.758 / 0.352

pH = 5.332 + 0.333

pH = 5.665

Question 2

Calculation of moles of NaA, HA and KOH.

We have relation, Molarity = No. of moles of solute / Volume of solution in L

No. of moles of solute = Molarity \times volume of solution in L

No. of moles of HA = 0.352 mol / L \times 1.00 L = 0.352 mol

No. of moles of NaA = 0.758 mol / L \times 1.00 L = 0.758 mol

No. of moles of KOH = 0.138 mol / L \times 0.310 L = 0.04278 mol

Consider reaction of KOH with buffer solution.

HA + OH -  \rightarrow A - + H2O

Let's use ICE table.

Moles HA OH - A -
I 0.352 0.04278 0.758
C -0.04278 -0.04278 +0.04278
E 0.30922 0.00000 0.80078

Volume of buffer solution after addition of KOH = 1.00 L + 0.310 L = 1.310 L

We have equation, pH = - log Ka + log [ NaA ] / [ HA ]

\therefore pH = - log 4.66 \times 10 -06 + log ( 0.80078 mol / 1.310 L ) / ( 0.30922 mol / 1.310 L )

pH = 5.332 + 0.4132

pH =5.745

Question 3

Calculation of moles of NaA, HA and KOH.

We have relation, Molarity = No. of moles of solute / Volume of solution in L

No. of moles of solute = Molarity \times volume of solution in L

No. of moles of HA = 0.352 mol / L \times 1.00 L = 0.352 mol

No. of moles of NaA = 0.758 mol / L \times 1.00 L = 0.758 mol

No. of moles of KOH = 0.138 mol / L \times 0.620 L = 0.08556 mol

Consider reaction of KOH with buffer solution.

HA + OH -  \rightarrow A - + H2O

Let's use ICE table.

Moles HA OH -   A -
I 0.352 0.08556 0.758
C -0.08556 -0.08556 +0.08556
E 0.26644 0.00000 0.84356

Volume of buffer solution after addition of KOH = 1.00 L + 0.620 L = 1.620 L

We have equation, pH = - log Ka + log [ NaA ] / [ HA ]

\therefore pH = - log 4.66 \times 10 -06 + log ( 0.84356 mol / 1.620 L ) / ( 0.26644 mol / 1.620 L )

pH = 5.332 + 0.5005

pH =5.832

Add a comment
Know the answer?
Add Answer to:
A 1.00 L solution contains 0.352 M HA (Ka = 4.66⋅10−64.66⋅10-6) and 0.758 M NaA. What...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT