Buffer solution Questions in English

(NONE) This is a basic buffer solution consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.
Given $pK_b$ for $NH_4OH$ is $4.74$.
The number of millimoles of base $(NH_4OH)$ = $5 \text{ mL} \times 0.1 \text{ M} = 0.5 \text{ mmol}$.
The number of millimoles of salt $(NH_4Cl)$ = $250 \text{ mL} \times 0.1 \text{ M} = 25 \text{ mmol}$.
The total volume of the solution = $5 \text{ mL} + 250 \text{ mL} = 255 \text{ mL}$.
Since the volume is the same for both, the ratio of concentrations is equal to the ratio of millimoles: $\frac{[Salt]}{[Base]} = \frac{25}{0.5} = 50$.
Now, $pOH = 4.74 + \log(50) = 4.74 + 1.699 \approx 6.44$.
Since $pH + pOH = 14$, we have $pH = 14 - 6.44 = 7.56$.
Thus, the $pH$ is $756 \times 10^{-2}$.

(B) For a buffer of weak acid $HX$ and its conjugate base $X^-$, the Henderson-Hasselbalch equation for $pOH$ is given by $pOH = pK_b + \log \frac{[X^-]}{[HX]}$.
Given that the concentrations are equal, $[X^-] = [HX]$, so $\log \frac{[X^-]}{[HX]} = \log(1) = 0$.
Therefore, $pOH = pK_b = -\log(K_b) = -\log(10^{-10}) = 10$.
Using the relation $pH + pOH = 14$ at $298 \text{ K}$, we get $pH = 14 - pOH = 14 - 10 = 4$.

(A) $1$. The mixture is a buffer solution consisting of a weak acid and its conjugate base.
$2$. Use the Henderson-Hasselbalch equation: $pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$.
$3$. Since equal volumes are mixed, the final concentration of both components is halved, but their ratio remains $\frac{0.1}{0.1} = 1$.
$4$. $pK_a = -\log(K_a) = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$.
$5$. $pH = 4.8861 + \log(1) = 4.8861 \approx 4.89$.

(A) Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First, calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
Since $[Salt] = [CH_3COONa] = 0.1 \text{ M}$ and $[Acid] = [CH_3COOH] = 0.1 \text{ M}$, the ratio $\frac{[Salt]}{[Acid]} = 1$.
Therefore, $pH = 4.745 + \log(1) = 4.745 + 0 = 4.745$.

(A) $1$. The solution is a basic buffer consisting of a weak base (pyridine) and its salt (pyridinium chloride).
$2$. The concentration of the base $[Base] = 0.2 \text{ M}$.
$3$. The concentration of the salt $[Salt] = \frac{0.15 \text{ mol}}{0.5 \text{ L}} = 0.3 \text{ M}$.
$4$. Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right)$.
$5$. $pOH = 8.75 + \log\left(\frac{0.3}{0.2}\right) = 8.75 + \log(1.5) = 8.75 + 0.176 = 8.926$.
$6$. Since $pH + pOH = 14$, $pH = 14 - 8.926 = 5.074 \approx 5.07$.