How would you calculate the pH of the buffer if 1.0mL of 5.0M NaOH is added to 20.0mL of this buffer?
Can someone please explain to me how to do B and C step by step so I could understand it clearly? :)
Say, for example, that you had prepared a buffer in which you mixed 10.30 g of sodium hydrogen phosphate, Na2HPO4, with 100.0 mL of 1.0 M sodium dihydrogen phosphate, NaH2PO4.
A. Calculate the pH of this buffer.
B. Calculate the resulting pH if 1.0 mL of 5.0 M NaOH is added to 20.0 mL of this buffer.
C. Calculate the mL of HCl that would have to be added to this buffer in order to lower the pH by exactly 1.0 unit.
Im not sure if I did A correctly, but...
Ka = 3.6x10^-13
pKa = -log(3.6x10,^-13)
= 12
10.30g Na2HPO4 x (1 mol Na2HPO4/ 141.96g Na2HPO4)
= 0.7256 mol Na2HPO4 (1.0 mol/1 L) x .1L
= .1 mol NaH2PO4
pH = 12 + log(.1 mol NaH2PO4/.07256 mol Na2HPO4)
= 12.14
For B. How exactly would I write the chemical equation and start the ICF table? And definitely need help on C. :(
Thank you soo much in advance.
No, A uses pK2 not pK3. And I think it is pK2 + log (BASE/ACID) but H2PO4 is the acid and HPO4 is the base. I calculate approx 8.1 but you should be more careful.
We'll take your values as ok.
0.7256 mol base
0.1 mol acid
So in 20 mL you will have 0.145 mol base and 0.02 mol acid.
For B. add 1mL of 5M NaOH = 0.005 mols
.........H2PO4^- + OH^- ==> HPO4^- + H2O
I.........0.02.......0....0.145
add..............0.005............
C......-0.005..-0.005.......+0.005
E.......0.015.......0........0.150
Substitute the E line into the HH equation and solve for pH.
C. I assume that is lowering of 1 pH unit for the solution in part A and not part B. pH lowered by 1 unit makes it about 7.1 from 8.1
..........HPO4^- + H^+ ==> H2PO4^-
I.......0.7256.....0........0.1
add HCl.............x.............
C..........-x......-x.......+x..
E........0.7256-x....0.......0.1+x
Substitute the E line into the HH equation and solve for x = mols HCl added. You want to calculate mL of ? M (you didn't give that along with not giving pK2.
7.1 = pK2 + log (0.7256-x)/(0.1+x)
and solve for x = mols HCl
Then M = mols HCl/L HCl
You know mols and M (not given in the post), solve for L and convert to mL.
Post your work if you get stuck. Check my work carefully, there are some contradictions in the problem and I may not have interpreted the problem correctly.
Oh right, the parent acid was H3PO4. In part A, when calculating the pH, what exactly is pKa2? I am still unsure how to find the Ka/Kb on the tables of Ka's and Kb's.
In part B, you mentioned that in 20 mL I will have 0.145 mol base and 0.02 mol acid.
The calculation for 0.2 mol acid is:
20 mL x (1 L / 1,000 mL x 1.0 mol / 1 L)
= 0.02 mol acid
right?
I am a bit confused how you got 0.145 mol base.
Is it:
.7256 mol base x .1 L = 7.256 M
7.256 M x.02 L
= 0.14512 mol base
Also, for 0.005 mol of NaOH, in which you got, would it not be 0.0005 mol NaOH?
Is it not 0.5 M x .001 L
= 0.00005 mol OH^-?
Thank you. :)
For part C, I just wanted to make sure if I did the calculations correctly when taking the anti-log.
7.07 = 7.21 + log (0.7256-x)/(0.1+x)
-0.14 = log (0.7256-x)/(0.1+x)
10^-0.14 = (0.7256-x)/(0.1+x)
(10^-0.14)(0.1+x) = (0.7256-x)
x = 0.9018 mol HCl
M of HCl. I'm assuming I am using the pH from part A? Although...this may be the wrong way to go about it...
[H+] = Ka * [HPO4-] / [H2PO4]
[H+] = 10^(-pH)
[H+] = 10^-8.07
[H+] = 8.51^-9 M
[HPO4-] = [H+] * [H2PO4] / Ka
= 8.51^-9 M * 1.0 M / 6.14x10^-8
= 0.1386 M HCl
I found out that Molarity of HCl was 0.5 M.
So then, 0.5 M HCl = 0.9018 mol HCl/ L HCl
L HCl = (0.9018 mol/0.5 M)
HCl = 1.8036 L x 1000 mL/1 L)
= 1803.6 mL
= 1800 mL HCl
To calculate the pH of the buffer, you need to consider the dissociation of the sodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) in water.
A. Calculation of the pH of the buffer:
1. Calculate the concentrations of Na2HPO4 and NaH2PO4:
- Na2HPO4: 10.30 g Na2HPO4 x (1 mol Na2HPO4/ 141.96 g Na2HPO4) = 0.07256 mol Na2HPO4
- NaH2PO4: 100.0 mL x (1 L/1000 mL) x 1.0 mol NaH2PO4/1 L = 0.100 mol NaH2PO4
2. Calculate the ratio of NaH2PO4 to Na2HPO4:
- [NaH2PO4] / [Na2HPO4] = 0.100 mol / 0.07256 mol = 1.378
3. Calculate the pKa:
- pKa = -log(Ka) = -log(3.6x10^(-13)) = 12
4. Calculate the pH of the buffer:
- pH = pKa + log([NaH2PO4] / [Na2HPO4]) = 12 + log(1.378) = 12.14 (correct calculation!)
Now let's move on to the next steps:
B. Calculation of the resulting pH if 1.0 mL of 5.0 M NaOH is added to 20.0 mL of this buffer:
1. Write the chemical equation for the reaction:
- NaOH + NaH2PO4 → Na2HPO4 + H2O
2. Determine the moles of NaOH added:
- Moles of NaOH = 1.0 mL x (1 L/1000 mL) x 5.0 mol/L = 0.005 mol
3. Determine the new concentrations of NaH2PO4 and Na2HPO4:
- NaH2PO4: 0.100 mol + 0.005 mol = 0.105 mol
- Na2HPO4: 0.07256 mol - 0.005 mol = 0.06756 mol
4. Calculate the new pH of the buffer using the same formula as step A.
C. Calculation of the mL of HCl needed to lower the pH by 1.0 unit:
1. Assume the buffer volume remains unchanged (100.0 mL).
2. Determine the new concentration of NaH2PO4 for the desired pH change:
- NaH2PO4 = (10^(-desired pH change)) x [Na2HPO4]
3. Determine the moles of NaH2PO4 needed:
- Moles of NaH2PO4 needed = (new concentration of NaH2PO4) x (buffer volume)
4. Calculate the mL of HCl needed based on the molarity of HCl solution used:
- mL of HCl = (moles of NaH2PO4 needed) / (molarity of HCl solution)
Remember to use the Henderson-Hasselbalch equation and appropriate logarithmic calculations for both parts B and C to ensure accurate and precise results.