How many milliliters of 1.55 M NaOH must be added to 125 mL of .19 M NaH2PO4 to make a buffer solution with a pH of 6.90?
I have attempted to solve this with no success, as this is an online homework assignment that requires an exact answer.
Hoping that you will be able to help. Thanks.
80.6 milliliters? O.o
I'm sorry but this answer didn't work...
......H2PO4^- + OH^- ==> HPO4^-2 + H2O
start..23.75 .....0.........0.........0
change..-x........x.........x.........x
end...23.75-x.....0........x...........x
Substitute this into the (B/A) = 0.5036 equation below.
[mmoles H2PO4 at start is from 125 mL x 0.19 M = 23.75 mmoles.
x = mmoles OH that must be added.]
What is the ratio of base to acid?
6.90 = pk2 + log[(base)/(acid)]
I looked up k2 for H3PO4 and found 6.34E-8 for pK2 = 7.2 but you need to use the pKa in your work, especially if your data base requires an exact answer. Also, you need to confirm all of my work and round to satisfy appropriate significant figures requirements. Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base)/(acid)
6.90 = 7.2 + log(B/A).
(base)/(acid) = 0.5036 or
base = 0.5036*acid
base is x from line 3 above. acid = 23.75-x.
x = 0.5036(23.75-x)
Solve for x.
I get something like 7.95 mmoles for the amount of NaOH to add.
How much 1.55 M NaOH is needed to give 0.00795 moles. M = moles/L. Solve for L and I get 0.00513 L or 5.13 mL.
Then you need to check this to make sure it produces a pH of 6.90.
5.13 mL x 1.55 M NaOH = 7.95 mmoles.
This will form 7.95 mmoles HPO4^-2(base)
This will leave 23.75-7.95 = 15.80 mmoles H2PO4^- (acid); then,
pH = 7.2 + log (7.95/15.8) = 6.90
Check my thinking. Check my work. Confirm all of this.
To solve this problem, we need to use the Henderson-Hasselbalch equation for buffer solutions. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of its acid and conjugate base:
pH = pKa + log([A-]/[HA])
where pH is the desired pH, pKa is the logarithmic acid dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, we are given the pH (6.90) and the concentrations of NaH2PO4 (0.19 M) and NaOH (1.55 M). The pKa for the H2PO4- ion is 7.21.
First, we need to determine the ratio of [A-] to [HA]. Since NaH2PO4 is the acid, [HA] = 0.19 M. To find [A-], we need to use the stoichiometry of the reaction:
NaH2PO4 + NaOH -> Na2HPO4 + H2O
The stoichiometry tells us that for every 1 mole of NaH2PO4 that reacts, we form 1 mole of Na2HPO4 and 1 mole of H2O. However, we have excess NaOH, so the limiting reagent is NaH2PO4. Therefore, the initial concentration of NaH2PO4 is the same as the concentration of H2PO4-, which is equal to [A-].
Now we can plug the values into the Henderson-Hasselbalch equation:
6.90 = 7.21 + log([A-]/0.19)
Rearranging the equation gives:
log([A-]/0.19) = -0.31
Next, we need to convert this logarithmic equation into an exponential one:
10^(-0.31) = [A-] / 0.19
Solving for [A-], we find:
[A-] = 0.19 * 10^(-0.31)
Now we know the concentration of the conjugate base, [A-], which is also the concentration of NaH2PO4. However, we need to find the volume of NaOH that needs to be added.
To determine the volume of NaOH, we can use the following equation:
C1V1 = C2V2
where C1 is the initial concentration of the solute (NaOH), V1 is the volume of the solute, C2 is the final concentration of the solute (which is 1.55 M), and V2 is the final volume of the solute (125 mL + V).
Plugging in the values, we get:
1.55 * V = 0.19 * 10^(-0.31) * (125 + V)
Solving for V, we find:
V = (0.19 * 10^(-0.31) * 125) / (1.55 - 0.19 * 10^(-0.31))
Now you can calculate the exact volume of NaOH that needs to be added to the solution to obtain a buffer solution with a pH of 6.90.