All categories

3 years ago

Need to complete the chart

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

Row 1

$[H^+]=1.8\times 10^{-6}M$

$pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7$

pOh=14-pH=14-5.7=8.3

$pOH=-\log[OH^-]$

$[OH^-]=0.5\times 10^{-8}M$

Hence, acidic

Row 2

$[OH^-]=3.6\times 10^{-10}M$

$pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4$

pH=14-pOH=14 - 9.4 = 4.6

$pH=-\log[H^+]$

$[H^+]=2.6\times 10^{-5}M$

Hence, acidic

Row 3

pH = 8.15

$[H^+]=0.7\times 10^{-8}M$

pOH=14-pH=14 - 8.15 = 5.8

$pOH=-\log[OH^-]$

$[OH^-]=1.5\times 10^{-6}M$

Hence, basic

Row 4

pOH = 5.70

$[OH^-]=1.8\times 10^{-6}M$

pH=14-pOH=14 - 5.70 = 8.3

$pH=-\log[H^+]$

$[H^+]=0.5\times 10^{-8}M$

Hence, basic

You might be interested in

Which of the following compounds contains both ionic and covalent bonds? KOH N2O5 CH3OH Na2O

dusya [7]

Answer: a.KOH

Potassium hydroxide is an ionic compound where the K+ is the cation and OH−is the anion. At the same time, the compound also contains a covalent bond since the anion, OH−is formed from electron sharing between the O and H atoms.

Hope this helps........ Stay safe and have a Merry Christmas! :D

Explanation:

4 0

2 years ago

Công thức của A có dạng Ca(hco3)x có ptk là 162 tìm x

harina [27]

Answer:

Explanation:

nwjwjw

dodkekds

dldldle

5 0

1 year ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,