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2 years ago

What is the ph of a 0.034 m hydrochloric acid solution?

Chemistry

2 answers:

Law Incorporation [45]2 years ago

8 0

PH= -log [H+]
= - (log 0.034)
= - (-1.5)
= 1.5

leva [86]2 years ago

4 0

Answer: The pH of the given HCl solution is 1.47

Explanation:

We are given:

Molarity of HCl = 0.034 M

1 mole of HCl produces 1 mole of hydrogen ions and 1 mole of chlorine ions.

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

Mathematically,

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

We are given:

$[H^+]=0.034M$

Putting values in above equation, we get:

$pH=-\log(0.034)\\\\pH=1.47$

Hence, the pH of the given HCl solution is 1.47

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3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

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