Chemistry, 26.01.2020 02:31 zaeswag23

Asolution has a [oh−] of 1 × 10−12 m. what is the [h3o+] of the solution?

a 1 × 10−14 m
b 1 × 10−2 m
c 2 m
d 14 m

Answers: 3

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Answer from: gianni0

1 × 10−2 M

Explanation: Kw is known as the constant of water dissociation and its value is found to be $10^{-14}$ . It is the product of the concentration of [H3O+] and [OH -] ions .

Dissociation of water can be represented by -

$2 H_{2} O (aq)\rightleftharpoons H_{3}O^{+}(aq) + OH^{-}(aq)$

Kw = [H3O+] [OH -]

As we know that pKw = - log Kw,

Thus pKw = 14

pOH = - log OH- = - log (10^{-12}) = 12

and pKw = pH + pOH

14 = pH + 12

pH = 14 - 12 = 2

pH = -log H3O+

2 = -log H3O+

Therefore, [H3O+] = 1 × 10−2 M

Answer from: natalyarenassalgado

The correct answer is Option B.

Explanation:

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

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

We are given:

$[OH^-]=1\times 10^{-12}M$

Putting values in above equation, we get:

$pOH=-\log(1\times 10^{-12})\\\\pOH=12$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-12=2$

pH is defined as negative logarithm of hydrogen ion or hydronium ion concentration. It is basically defined as the power of hydronium ions in a solution.

$pH=-\log[H_3O^+]$

pH of the solution = 2

Putting values in above equation, we get:

$2=-\log[H_3O^+]$

[H_3O^+]=antilog(-2)

$[H_3O^+]=1\times 10^{-2}$

Hence, the correct answer is Option B.

Answer from: Quest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

Answer from: Quest

it is stable as it is in group 0 which have full outer shells of electrons

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