Question

1. which of the following are buffer solutions? explain you reasoning.

a. 0.100 m koh
b. 0.100 m koh and 0.100 m hbr
c. 0.100 m (nh₄)₂so₄ and 0.100 m nh₃
d. 0.100 m hcl and 0.200 m nano₂

1. you are asked to prepare a buffer solution with a ph of 3.50. the following 0.100 m solutions are available for your use: hcooh, ch₃cooh, h₃po₄, nahcoo, nach₃coo, and nah₂po₄. describe how you would make the buffer (include volumes).
2. calculate the ratio of hpo₄²⁻ to h₂po₄⁻ required to maintain a ph of 7.1.

Explanation:

Step1: Recall buffer solution criteria

A buffer solution is a mixture of a weak acid - conjugate base pair or a weak base - conjugate acid pair.

Step2: Analyze option a

$0.100\ M\ KOH$ is a strong base. It cannot form a buffer on its own as it lacks a weak acid - conjugate base or weak base - conjugate acid pair. So it is not a buffer.

Step3: Analyze option b

$KOH$ is a strong base and $HBr$ is a strong acid. Their reaction will result in a neutralization reaction and not form a buffer as there are no weak acid - conjugate base or weak base - conjugate acid components left in the right proportions for buffering action. So it is not a buffer.

Step4: Analyze option c

$(NH_4)_2SO_4$ provides $NH_4^+$ (the conjugate acid) and $NH_3$ is the weak base. This is a weak base - conjugate acid pair. So it is a buffer.

Step5: Analyze option d

$HCl$ is a strong acid. But when it reacts with $NaNO_2$, it forms $HNO_2$ (a weak acid) and $NaCl$. The remaining $NaNO_2$ and the formed $HNO_2$ create a weak acid - conjugate base pair. So it is a buffer.

For question 5:
We use the Henderson - Hasselbalch equation $pH = pK_a+\log\frac{[A^-]}{[HA]}$.
For $HCOOH$, $pK_a = -\log(1.8\times 10^{-4})\approx3.74$.
Let $V_1$ be the volume of $HCOOH$ and $V_2$ be the volume of $NaHCOO$.
We want $pH = 3.50$. So $3.50=3.74+\log\frac{V_2}{V_1}$
$\log\frac{V_2}{V_1}=3.50 - 3.74=- 0.24$
$\frac{V_2}{V_1}=10^{- 0.24}\approx0.575$.
Let the total volume $V = V_1 + V_2$. If we assume $V = 1\ L$, then $V_2=0.575V_1$. Also $V_1+V_2 = 1\ L$, so $V_1+0.575V_1 = 1\ L$, $V_1=\frac{1}{1 + 0.575}\approx0.635\ L$ and $V_2=1 - 0.635 = 0.365\ L$.

For question 6:
For the $H_2PO_4^--HPO_4^{2 - }$ buffer system, $pK_{a2}$ of $H_3PO_4$ is $7.21$.
Using the Henderson - Hasselbalch equation $pH = pK_a+\log\frac{[A^-]}{[HA]}$, where $A^-$ is $HPO_4^{2 - }$ and $HA$ is $H_2PO_4^-$.
Given $pH = 7.1$, $7.1=7.21+\log\frac{[HPO_4^{2 - }]}{[H_2PO_4^-]}$
$\log\frac{[HPO_4^{2 - }]}{[H_2PO_4^-]}=7.1 - 7.21=-0.11$
$\frac{[HPO_4^{2 - }]}{[H_2PO_4^-]}=10^{-0.11}\approx0.776$.

Answer:

a. Not a buffer.
b. Not a buffer.
c. Buffer.
d. Buffer.
For question 5: Mix approximately $0.635\ L$ of $0.100\ M\ HCOOH$ and $0.365\ L$ of $0.100\ M\ NaHCOO$ to make the buffer.
For question 6: The ratio $\frac{[HPO_4^{2 - }]}{[H_2PO_4^-]}$ is approximately $0.776$.