Question

2. A 350 mL buffer solution containing 0.15 M HF and 0.150 M NaF is reacted with sodium hydroxide. What mass of NaOH can this buffer neutralize before the pH rises above 4.0? If the same volume of buffer was 0.35 M in HF and 0.35 M in NaF then what mass of NaOH could be handled before the pH rises above 4.0?

Answer 1

First calculate pH of the buffer.
pH = 3.14 + log(0.15/0.15)
pH = 3.14 based on Ka for HF of 7.2E-4. You need to use the K in your text and/or notes.
Then substitute 4.0 for pH and solve for (F^-/HF). If I didn't goof I found 7.24.
Here is the equation that goes when adding NaOH.
......HF...+....OH^- ===>F^- + H2O
begin 0.0525....0........0.0525 moles
add OH..........x..........x
final.0.525-x...0.......0.0525+x....
Now substitute into the (F^-/HF) = 7.24 and solve for x. x will be the moles of NaOH that can be added without exceeding a pH of 4.00. Then convert that many moles to grams NaOH.
The second part of the problem is done the same way.

Answer 2

thanks for the explanation!!!

Answer 3

1+1=3



:)