What volume of 0.100 M NaOH is needed to make 100.0 mL of a buffer solution with a pH of 6.00 if one starts with 50.0 mL of 0.100 M potassium hydrogen phthalate? The Ka2 for potassium hydrogen phthalate is 3.1 × 10-6.
To find the volume of NaOH needed to make a buffer solution with a pH of 6.00, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this case, the acid (HA) is potassium hydrogen phthalate (KHP), and the conjugate base (A-) is the resulting salt after adding NaOH.
Given information:
- pH = 6.00
- pKa = -log(Ka) = -log(3.1 × 10-6)
Now, let's solve for [A-]/[HA]:
6.00 = pKa + log([A-]/[HA])
6.00 = -log(3.1 × 10-6) + log([A-]/[HA])
6.00 + log(3.1 × 10-6) = log([A-]/[HA])
log([A-]/[HA]) = 6.00 + log(3.1 × 10-6)
Now, to find [A-]/[HA], we need to convert the pH back to [H+] concentration:
[H+] = 10^(-pH)
[H+] = 10^(-6.00)
Next, we can use the stoichiometry of the reaction between KHP and NaOH to find the moles of NaOH needed for the reaction:
moles of NaOH = moles of KHP
Moles of KHP can be calculated using the concentration (M) and volume (L) of KHP:
moles of KHP = concentration (M) * volume (L)
moles of KHP = 0.100 M * 0.0500 L
Now we can use the balanced equation between KHP and NaOH to determine the stoichiometry:
KHP + NaOH -> KNaP + H2O
From the balanced equation, we can see that 1 mole of KHP reacts with 1 mole of NaOH to form 1 mole of KNaP. Therefore, moles of NaOH needed = moles of KHP.
Now, using moles of NaOH, we can calculate the volume of 0.100 M NaOH needed:
volume (L) of NaOH = moles of NaOH / concentration (M) of NaOH
volume (L) of NaOH = moles of KHP / concentration (M) of NaOH
Finally, substitute the values and calculate:
[H+] = 10^(-6.00) = 1.0 x 10^(-6) M
moles of KHP = 0.100 M * 0.0500 L = 0.005 moles
volume (L) of NaOH = 0.005 moles / 0.100 M = 0.050 L or 50.0 mL
Therefore, the volume of 0.100 M NaOH needed to make 100.0 mL of a buffer solution with a pH of 6.00, starting with 50.0 mL of 0.100 M KHP, is also 50.0 mL.