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3 years ago

What are the pH of these solutions?

Chemistry

1 answer:

Tju [1.3M]3 years ago

3 0

Answer:

The answer to your question is below

Explanation:

a) HCl 0.01 M

pH = -log [0.01]

pH = - (-2)

pH = 2

b) HCl = 0.001 M

pH = -log[0.001]

pH = -(-3)

pH = 3

c) HCl = 0.00001 M

pH = -log[0.00001]

pH = - (-5)

pH = 5

d) Distilled water

pH = 7.0

e) NaOH = 0.00001 M

pOH = -log [0.00001]

pOH = -(-5)

pH = 14 - 5

pH = 9

f) NaOH = 0.001 M

pOH =- log [0.001]

pOH = 3

pH = 14 - 3

pH = 11

g) NaOH = 0.1 M

pOH = -log[0.1]

pOH = 1

pH = 14 - 1

pH = 13

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3 years ago

Calculate the molarity of the sodium acetate solution as described below.

Airida [17]

Answer:

This question is incomplete.

Explanation:

This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;

Molarity = number of moles (n) / volume (in liter or dm³)

To calculate the number of moles of NaC₂H₃O₂, we say

number of moles (n) =

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The unit for Molarity is M (Molar concentration), mol/L or mol/dm³

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2 years ago

What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?

AlexFokin [52]

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

$M=\frac{mol\ solute}{Volume\ solution \ in \ L}$

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

$mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L$