Question

Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4 · H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M

Answer 1

Answer:

• Mass of NaH₂PO₄·H₂O = 8.542 g
• Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + $log\frac{[A^{-}]}{[HA]}$

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can express [H₂PO₄⁻] in terms of [HPO₄⁻²]:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, in order to calculate [HPO₄⁻²]:

$7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]$

With the value of  [H₂PO₄⁻], we calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:

• Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g

• Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g