Question

0.1M Na2HPO4 and 0.2MNaH2PO4.How much stock solutions and H2O would be needed to prepare 0.5M,2L of phosphate buffer at pH7.4?(pKa fr H2PO4 :7.2) can you guide me through it? i don't understand how this should be done.thanks

Answer 1

Answer on Question#39444 - Chemistry - Inorganic Chemistry

Question

0.1M Na₂HPO₄ and 0.2M NaH₂PO₄. How much stock solutions and H₂O would be needed to prepare 0.5M, 2L of phosphate buffer at pH 7.4? (pKa fr H₂PO₄ : 7.2) can you guide me through it? I don't understand how this should be done. Thanks

Answer:

The pH of buffer solutions can be determined using the following equation:

$$\mathrm{pH} = \mathrm{pK_a} + \lg [\mathrm{B'}] / [\mathrm{HB}]$$

where [HB] is the concentration of acid, and [B'] is the base concentration. In case of discussed buffer, the acid is $\mathrm{NaH_2PO_4}$, and the base is $\mathrm{Na_2HPO_4}$. Hence, knowing the pH and $\mathrm{pK_a}$ values, we can obtain the molar ratio between acidic and basic compounds:

$$\lg [\mathrm{Na_2HPO_4}] / [\mathrm{NaH_2PO_4}] = \mathrm{pH} - \mathrm{pK_a}$$ $$[\mathrm{Na_2HPO_4}] / [\mathrm{NaH_2PO_4}] = 1.58$$

Dilution does not affect on pH of the buffer, hence we can simply compare the amounts of $\mathrm{NaH_2PO_4}$ and $\mathrm{Na_2HPO_4}$ in the final solution.

Let "a" denote the volume of $\mathrm{NaH_2PO_4}$ and "b" denote the volume of $\mathrm{Na_2HPO_4}$.

$$n(\mathrm{NaH_2PO_4}) = a \cdot 0.2; \quad n(\mathrm{Na_2HPO_4}) = b \cdot 0.1$$ $$n(\mathrm{Na_2HPO_4}) / n(\mathrm{NaH_2PO_4}) = 1.58$$ $$a/b = 3.16$$

So, to obtain buffer with $\mathrm{pH} = 7.4$ you should take 3.16 parts of 0.1M $\mathrm{Na_2HPO_4}$ per 1 part of 0.2M $\mathrm{NaH_2PO_4}$.