You have 800 mL of a 0.050M phosphate buffer, pH 6.5. You need to increase the pH of this buffer to 7.5 by using 6.00M NaOH. Determine the volume of NaOH needed. What will be the final concentration of the buffer?
I do these this way.
I used 6.34E-8 for k2.
6.5 = 7.2 + log (base)/(acid)
That gives me (base)/(acid) = 0.2 equation 1.
Equation 2 is acid + base = 0.05M
Solve these two equations simultaneously to obtain (acid) = 0.0417M
(base) = 0.00833M.
Then I change these to millimols.
800 x 0.0417 = 33.36 mmoles acid.
800 x 0.0833 = 6.667 mmols base
You want to add NaOH to make it 7.5
.........acid + OH^- ==> base + H2O
I........33.36....0.......6.667
.add.............x..............
C.........-x.....-x.........x
E.......33.36-x...x.........x
Substitute into the HH equation to obtain
7.5 = 7.2 + log base/acid
7.5 = 7.2 + log (6.667+x)/(33.36-x)
Solve for x = mmoles OH^- needed.
Since M = mmols/mL, plug in to find mL OH^- needed.
Post your work if you get stuck.
To determine the volume of NaOH needed to increase the pH of the buffer, we need to use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where pH is the desired pH, pKa is the acid dissociation constant of the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In this case, the buffer is made up of the acid H2PO4- (concentration [HA]) and its conjugate base HPO4^2- (concentration [A-]).
First, we need to find the initial concentrations of [HA] and [A-].
Given:
Initial volume of the buffer solution = 800 mL
Initial concentration of the buffer solution = 0.050 M
Since the initial concentration of the buffer solution is 0.050 M, the concentrations of [HA] and [A-] are the same. Therefore, [HA] = [A-] = 0.050 M.
Now, we want to increase the pH of the buffer from 6.5 to 7.5 using 6.00 M NaOH.
To find the volume of NaOH needed, we can use the equation:
[V1 * C1] + [V2 * C2] = [V1 + V2] * C_final
Where V1 is the initial volume of the buffer, C1 is the initial concentration of the buffer, V2 is the volume of NaOH added, C2 is the concentration of NaOH, and C_final is the final concentration of the buffer.
Rearranging the equation to solve for V2:
V2 = ([V1 + V2] * C_final - V1 * C1) / C2
Plugging in the given values:
V1 = 800 mL = 0.800 L
C1 = 0.050 M
C_final = 0.050 M (The final concentration of the buffer remains the same)
To convert the concentration of NaOH to molar (M) units, we need to multiply it by its stoichiometric factor. In this case, the stoichiometric factor is 1 as the balanced equation between NaOH and HPO4^2- is:
NaOH + HPO4^2- → NaHPO4
Thus, 6.00 M NaOH remains the same.
Substituting the values into the equation:
V2 = ([0.800 L + V2] * 0.050 M - 0.800 L * 0.050 M) / 6.00 M
Simplifying:
V2 = (0.040 L - 0.040 L) / 6.00 M
V2 = 0 L / 6.00 M
Therefore, the volume of NaOH needed to increase the pH of the buffer is 0 L.
As for the final concentration of the buffer, it remains the same at 0.050 M as no additional water or buffer components were added or removed.