Question

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

0.40 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?

0.15 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 5.5 L of solution. What is the pH of this solution?

Answer 1

1)

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 0.40 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(0.4 g)/(36.458 g/mol)

= 1.097*10^-2 mol

volume , V = 8.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 1.097*10^-2/8

= 1.371*10^-3 M

So,

[H+] = 1.371*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (1.371*10^-3)

= 2.86

Answer: 2.86

2)

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH = 0.15 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(0.15 g)/(39.998 g/mol)

= 3.75*10^-3 mol

volume , V = 5.5 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 3.75*10^-3/5.5

= 6.819*10^-4 M

So,

[OH-] = 6.819*10^-4 M

we have below equation to be used:

pOH = -log [OH-]

= -log (6.819*10^-4)

= 3.17

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.17

= 10.83

Answer: 10.83