Question

Consider a solution prepared by mixing the following:50.0 mL of 0.100 M Na3PO4100.0 mL of 0.0500 M KOH200.0 mL of 0.0750 M HCl50.0 mL of 0.150 M NaCNDetermine the volume of 0.100 M HNO3 that must be added to this mixture to achieve a nal pH value of 7.21.

Answer 1

Answer:

you must add 50 mL

Explanation:

Hi

KOH is a strong base and by adding 100mL 0.05M you will have an amount of 5 millimol.

NaCN is a base and by adding 50 mL 0,150 M you will have an amount of 7,5 mmol.

HCl is a acid and by adding 200 mL 0,075 M you will have an amount of 15 mmol.

The acid reacts with the bases leaving 2.5 mmol unreacted.

Na3PO4 is a base and by adding 50 mL 0,1 M you will have an amount of 5 mmol.

The 2.5 mmol of acid react with the base PO4 ^ -3 forming a regulatory solution of PO4 ^ -3 and HPO4 ^ -2 of pKa 2.12

5 mmol of acid (HNO3) must be added to obtain a regulatory solution formed by the same amount of HPO4 ^ -2 (2.5 mmol) and H2PO4 ^ -1 (2.5 mmol) with pKa 7.21

Considering a quantity of 5 mmol of HNO3 of concentration 0.1 M, 50 mL must be added.

To calculate the pH of the regulatory solution you should consider pH = pKa × Ca / Cb pH = 7.21 × 2.5 / 2.5 = 7.21 Being in the same solution the volume is the same and can be simplified to achieve a faster calculation.

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