question 6 of 8\nphosphoric acid is a triprotic acid with the pkₐ values pkₐ₁ = 2.148, pkₐ₂ = 7.198, and…

question 6 of 8\nphosphoric acid is a triprotic acid with the pkₐ values pkₐ₁ = 2.148, pkₐ₂ = 7.198, and pkₐ₃ = 12.375. you wish to prepare 1.000 l of a 0.0100 m phosphate buffer at ph 7.500. to do this, you choose to mix the two salt forms involved in the second ionization, nah₂po₄ and na₂hpo₄, in a 1.000 l volumetric flask and add water to the mark. what mass of each salt will you add to the mixture?\nmass nah₂po₄ = 0.399 g\nmass na₂hpo₄ = 0.947 g\nwhat other combination of phosphoric acid and its salts could be mixed to prepare this buffer?\n□ na₂hpo₄ and na₃po₄\n□ h₃po₄ and na₃po₄\n□ h₃po₄ and na₂hpo₄\n□ nah₂po₄ and na₃po₄\n□ h₃po₄ and nah₂po₄

question 6 of 8\nphosphoric acid is a triprotic acid with the pkₐ values pkₐ₁ = 2.148, pkₐ₂ = 7.198, and pkₐ₃ = 12.375. you wish to prepare 1.000 l of a 0.0100 m phosphate buffer at ph 7.500. to do this, you choose to mix the two salt forms involved in the second ionization, nah₂po₄ and na₂hpo₄, in a 1.000 l volumetric flask and add water to the mark. what mass of each salt will you add to the mixture?\nmass nah₂po₄ = 0.399 g\nmass na₂hpo₄ = 0.947 g\nwhat other combination of phosphoric acid and its salts could be mixed to prepare this buffer?\n□ na₂hpo₄ and na₃po₄\n□ h₃po₄ and na₃po₄\n□ h₃po₄ and na₂hpo₄\n□ nah₂po₄ and na₃po₄\n□ h₃po₄ and nah₂po₄

Answer

Explanation:

Step1: Use Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation for the second - ionization of phosphoric acid is $pH = pK_{a2}+\log\frac{[A^-]}{[HA]}$, where $pK_{a2} = 7.198$, $pH = 7.500$, $[A^-]=[Na_2HPO_4]$ and $[HA]=[NaH_2PO_4]$. $7.500=7.198+\log\frac{[Na_2HPO_4]}{[NaH_2PO_4]}$ $\log\frac{[Na_2HPO_4]}{[NaH_2PO_4]}=7.500 - 7.198=0.302$ $\frac{[Na_2HPO_4]}{[NaH_2PO_4]} = 10^{0.302}\approx2.00$ Also, $[Na_2HPO_4]+[NaH_2PO_4]=0.0100\ M$ (total buffer concentration) Let $x = [NaH_2PO_4]$, then $[Na_2HPO_4]=2x$ $x + 2x=0.0100\ M$, so $3x = 0.0100\ M$ and $x=[NaH_2PO_4]=\frac{0.0100}{3}\ M$, $[Na_2HPO_4]=\frac{2\times0.0100}{3}\ M$

Step2: Calculate moles in 1.000 L

Since $n = C\times V$ and $V = 1.000\ L$, $n(NaH_2PO_4)=[NaH_2PO_4]\times V=\frac{0.0100}{3}\ mol$ and $n(Na_2HPO_4)=\frac{2\times0.0100}{3}\ mol$

Step3: Calculate masses

The molar mass of $NaH_2PO_4$ is $M(NaH_2PO_4)=22.99 + 2\times1.008+30.97+4\times16.00=119.98\ g/mol$ $m(NaH_2PO_4)=n(NaH_2PO_4)\times M(NaH_2PO_4)=\frac{0.0100}{3}\ mol\times119.98\ g/mol\approx0.399\ g$ The molar mass of $Na_2HPO_4$ is $M(Na_2HPO_4)=2\times22.99+1.008 + 30.97+4\times16.00 = 141.96\ g/mol$ $m(Na_2HPO_4)=n(Na_2HPO_4)\times M(Na_2HPO_4)=\frac{2\times0.0100}{3}\ mol\times141.96\ g/mol\approx0.947\ g$

For the second part: A buffer is most effective when $pH$ is within $\pm1$ of $pK_{a}$ value. Since $pH = 7.500$ and $pK_{a2}=7.198$, we need a conjugate - acid base pair centered around $pK_{a2}$. A buffer can be made from a weak acid and its conjugate base. The conjugate - acid base pair for the second ionization step is $H_3PO_4$ and $Na_2HPO_4$.

Answer:

mass $NaH_2PO_4 = 0.399\ g$ mass $Na_2HPO_4 = 0.947\ g$ $H_3PO_4$ and $Na_2HPO_4$