PART A: 0.10 g of hydrogen chloride (HCl ) is dissolved in water to make 8.0 L o

Question

PART A: 0.10 g of hydrogen chloride (HCl ) is dissolved in water to make 8.0 L of solution. What is the pH of the resulting hydrochloric acid solution?

PART B: 0.60 g of sodium hydroxide (NaOH ) pellets are dissolved in water to make 2.0 L of solution. What is the pH of this solution?

PART C: Calculate the concentration of H+ ions in a 0.010 M aqueous solution of sulfuric acid.

PART D: A certain weak acid, HA , has a Ka value of 7.8×107. Calculate the percent ionization of HA in a 0.10 M solution.

PLEASE EXPLAIN YOUR STEPS and the logic behind if possible. Thank you!

Explanation / Answer

A)

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.5

= 36.5 g/mol

m(HCl)= 0.10 g

number of mol of HCl,

n = mass/molar mass

=(0.1 g)/(36.5 g/mol)

= 2.74*10^-3 mol

volume , V = 8.0 L

Molarity,

M = number of mol / volume in L

= 2.74*10^-3 mol /8 L

= 3.425*10^-4 M

So,

[H+] = 3.425*10^-4 M

now use:

pH = -log [H+]

pH = -log (3.425*10^-4)

pH = 3.5

Answer: 3.5

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