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 *K*a value of
7.8×10^{−7}. Calculate the percent ionization of HA in a
0.10 *M* solution.

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

Answer #1

**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**

**Rest of the questions are not related to this
question**

**They will be considered as different
questions**

**I am allowed to answer only 1 question at a
time**

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The temperature for each solution is carried out at
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