Question

Part C

A 0.161 M weak acid solution has a pH of 4.29. FindKa for the acid.

Express your answer using two significant figures.

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

Part D

0.30 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution?

Express the pH numerically to two decimal places.

Part E

0.80 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 3.0 L of solution. What is the pH of this solution?

Express the pH numerically to two decimal places

Answer 1

C)

we have below equation to be used:

pH = -log [H+]

4.29 = -log [H+]

log [H+] = -4.29

[H+] = 10^(-4.29)

[H+] = 5.129*10^-5 M

Lets write the dissociation equation of HA

HA -----> H+ + A-

0.161 0 0

0.161-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 5.129*10^-5*5.129*10^-5/(0.161-5.129*10^-5)

Ka = 1.634*10^-8

Answer: Ka = 1.6*10^-8

D)

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

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 0.30 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(0.3 g)/(36.458 g/mol)

= 8.229*10^-3 mol

volume , V = 4.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 8.229*10^-3/4

= 2.057*10^-3 M

So,

[H+]= 2.057*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.057*10^-3)

= 2.69

Answer: 2.69

E)

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 = 0.80 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(0.8 g)/(39.998 g/mol)

= 2*10^-2 mol

volume , V = 3.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2*10^-2/3

= 6.667*10^-3 M

So,

[OH-] = 6.667*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (6.667*10^-3)

= 2.1761

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.1761

= 11.82

Answer: 11.82