Calculate the volume of NaOH 0.1M that needs to be calculated to be added to a 30ml o-phosphoric acid 0.1 in order to prepare solutions of pH 7.5 and pH 6.9? pKa3= 2.15; pKa2=7.2; pKa1 = 12.1
You have the pKa values wrong. pK1 = 2.15; pK2 = 7.2 (which you have right) and pK3 = 12.1.
Use the Henderson-Hasselbalch equation.
pH = pk2 + log (base)/(acid)
7.5 = 7.2 + log (base)/(acid) or
7.5 = 7.2 + log b/a
H3PO4 + NaOH ==> NaH2PO4 + H2O
NaH2PO4 + NaOH ==> Na2HPO4 + H2O
Na2HPO4 + NaOH ==> Na3PO4 + H2O
The one you want is the middle one since that's the closest to your desired pH of 7.5. millimols H2PO4^- you have is 30 mL x 0.1M = 3.
.....H2PO4^- + OH^- ==> HPO4^2- + H2O
I.....3........0..........0........0
add............x...................
C.....-x......-x...........x
E.....3-x......0...........x
Substitute the E line into the HH equation and solve for x = millimols OH to be added to H2PO4^- to obtain the pH you want. Then convert that many millimols NaOH to mL using the M of the NaOH. This is not the end of the problem. Then you must add mL of 0.1M NaOH needed to neutralize the first H^+ from H3PO4. That is, it takes 30 mL of 0.1M NaOH to neutralize H3PO4 to H2PO4^- and that's where your starting with this problem so that 30 mL must added to whatever you calculate for the pH you want. Post your work if you get stuck. I don't think I've missed anything.
Thanks for d reply. Pls if i want to calculate for a pH less than 7 (eg 5.8), wil i subtract d volume i obtain from 30ml or will i add it?
To calculate the volume of NaOH 0.1M required to prepare solutions of pH 7.5 and pH 6.9, we need to consider the acid-base equilibrium between o-phosphoric acid (H3PO4) and NaOH. The pKa values provided are the dissociation constants for the three ionizable protons in phosphoric acid.
First, let's determine the equation for the deprotonation of o-phosphoric acid:
H3PO4 (aq) ⇌ H2PO4- (aq) + H+ (aq) (Equation 1)
H2PO4- (aq) ⇌ HPO42- (aq) + H+ (aq) (Equation 2)
HPO42- (aq) ⇌ PO43- (aq) + H+ (aq) (Equation 3)
Now, let's analyze the pH requirements:
1. For pH 7.5:
Since pH 7.5 is close to the second pKa (pKa2=7.2), we can assume that the concentration of H2PO4- (aq) and HPO42- (aq) are equal. Therefore, we only need to focus on Equation 2:
H2PO4- (aq) ⇌ HPO42- (aq) + H+ (aq)
2. For pH 6.9:
Since pH 6.9 is below pKa2 (7.2), we need to consider Equation 1, Equation 2, and Equation 3:
H3PO4 (aq) ⇌ H2PO4- (aq) + H+ (aq)
H2PO4- (aq) ⇌ HPO42- (aq) + H+ (aq)
HPO42- (aq) ⇌ PO43- (aq) + H+ (aq)
Now, let's calculate the moles of H+ ions required to achieve pH 7.5 and pH 6.9 in the 30 mL solution of o-phosphoric acid 0.1M. To do this, we need to use the Henderson-Hasselbalch equation:
For pH 7.5:
pH = pKa + log ([A-]/[HA])
7.5 = 7.2 + log ([HPO42-]/[H2PO4-])
0.3 = log ([HPO42-]/[H2PO4-])
[HPO42-]/[H2PO4-] = 10^0.3
[HPO42-]/[H2PO4-] = 2
For pH 6.9:
pH = pKa + log ([A-]/[HA])
6.9 = 7.2 + log ([HPO42-]/[H2PO4-])
-0.3 = log ([HPO42-]/[H2PO4-])
[HPO42-]/[H2PO4-] = 10^-0.3
[HPO42-]/[H2PO4-] = 0.5012
Using these ratios, we can calculate the moles of H+ ions required to adjust the pH:
For pH 7.5:
[HPO42-]/[H2PO4-] = 2
[HPO42-] = 2[H2PO4-]
[HPO42-] = 2 * (0.1 M) * (30 mL) = 0.6 mmol
For pH 6.9:
[HPO42-]/[H2PO4-] = 0.5012
[HPO42-] = 0.5012[H2PO4-]
[HPO42-] = 0.5012 * (0.1 M) * (30 mL) = 0.1504 mmol
Now, let's calculate the moles of NaOH required to neutralize the H+ ions:
For pH 7.5:
Since NaOH reacts in a 1:1 stoichiometry with H+, we need 0.6 mmol of NaOH.
For pH 6.9:
Since NaOH reacts in a 1:1 stoichiometry with H+, we need 0.1504 mmol of NaOH.
Now, let's calculate the volume of NaOH 0.1M required to obtain these moles:
For pH 7.5:
Volume of NaOH (0.1 M) = (0.6 mmol) / (0.1 M) = 6 mL
For pH 6.9:
Volume of NaOH (0.1 M) = (0.1504 mmol) / (0.1 M) = 1.504 mL
Therefore, to prepare a solution of pH 7.5, you need to add 6 mL of NaOH 0.1M to the 30 mL o-phosphoric acid 0.1M. And to prepare a solution of pH 6.9, you need to add 1.504 mL of NaOH 0.1M to the 30 mL o-phosphoric acid 0.1M.