Question

The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10−14

Part A

0.25 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

0.90 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 6.5 L of solution. What is the pH of this solution?

Express the pH numerically to two decimal places

part C

What is the pOH of the solution in part B ?.

.

Answer 1

Part A :

Number of moles of HCl , n = mass/molar mass

= 0.25 g / ( 36.5(g/mol)

= 6.85x10^-3 mol

So concentration of HCl = number of moles / volume in L

= 6.85x10^-3 mol / 8.0 L

= 8.56x10^-4 M

HCl ---> H⁺ + Cl^-

1 mole of HCl contains 1 mole of H⁺

So [H⁺] = [HCl] = 8.56x10^-4 M

pH = - log[H⁺]

= - log(8.56x10^-4 )

= 3.07

Part B :

Molarity of NaOH = ( mass/molar mass) / volume in L

= ( 0.90 g / 40(g/mol) ) / 6.5 L

= 3.46x10^-3 M

1 mole of NaOH contains 1 mole of OH-

[OH^-] = [NaOH] = 3.46x10^-3 M

pOH = - log (3.46x10^-3 )

= 2.46

pH = 14-pOH

= 14-2.46

= 11.54

Part C :

pOH = 2.46