Evaluate 15/k when k=3

Mathematics

2 answers:

umka21 [38]2 years ago

8 0

Answer:

Step-by-step explanation:

Hope this helps!

iVinArrow [24]2 years ago

7 0

Answer:

Step-by-step explanation:

K equals 3 and so 15/3 equals 5.

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Calculate the pH of a buffer solution made by mixing 300 mL of 0.2 M acetic acid, CH3COOH, and 200 mL of 0.3 M of its salt sodiu

Lesechka [4]

Answer:

Approximately $4.75$ .

Step-by-step explanation:

Remark: this approach make use of the fact that in the original solution, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ are equal.

${\rm CH_3COOH} \rightleftharpoons {\rm CH_3COO^{-}} + {\rm H^{+}}$

Since $\rm CH_3COONa$ is a salt soluble in water. Once in water, it would readily ionize to give $\rm CH_3COO^{-}$ and $\rm Na^{+}$ ions.

Assume that the $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ ions in this solution did not disintegrate at all. The solution would contain:

$0.3\; \rm L \times 0.2\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COOH$ , and

$0.06\; \rm mol$ of $\rm CH_3COO^{-}$ from $0.2\; \rm L \times 0.3\; \rm mol \cdot L^{-1} = 0.06\; \rm mol$ of $\rm CH_3COONa$ .

Accordingly, the concentration of $\rm CH_3COOH$ and $\rm CH_3COO^{-}$ would be:

$\begin{aligned} & c({\rm CH_3COOH}) \\ &= \frac{n({\rm CH_3COOH})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

$\begin{aligned} & c({\rm CH_3COO^{-}}) \\ &= \frac{n({\rm CH_3COO^{-}})}{V} \\ &= \frac{0.06\; \rm mol}{0.5\; \rm L} = 0.12\; \rm mol \cdot L^{-1} \end{aligned}$ .

In other words, in this buffer solution, the initial concentration of the weak acid $\rm CH_3COOH$ is the same as that of its conjugate base, $\rm CH_3COO^{-}$ .

Hence, once in equilibrium, the $\rm pH$ of this buffer solution would be the same as the ${\rm pK}_{a}$ of $\rm CH_3COOH$ .

Calculate the ${\rm pK}_{a}$ of $\rm CH_3COOH$ from its ${\rm K}_{a}$ :

$\begin{aligned} & {\rm pH}(\text{solution}) \\ &= {\rm pK}_{a} \\ &= -\log_{10}({\rm K}_{a}) \\ &= -\log_{10} (1.76 \times 10^{-5}) \\ &\approx 4.75\end{aligned}$ .