Question

Can someone please help me solve these, please show work/explanation and answer! I only need help with d and e for number 7. Then I do not know how to do 14 and 15.

7. Determine the pH of the following solutions 0.040 M HCI а. www b. 0.040 M КОН 0.0087 М НСIO С. 0.010 M CH3NH2 (methylamine) d. 0.018 M NaF е.

Answer 1

7.

d)

CH₃NH₂ is weak base, pKb of methylamine = 3.38

for weak base

pH = 14 - $\small \frac{1}{2}$ ( pKb - log C)

or , pH = 14 - $\small \frac{1}{2}$ ( 3.38 -  log 0.010)

or, pH = 14 - $\small \frac{1}{2}$ ( 3.38 + 2)

or, pH = 11.31

e)

NaF is a salt of weak acid HF and strong base NaOH

hence

pH = 7 + $\small \frac{1}{2}$ ( pKa + log C_salt)

pKa of HF = 3.1 M

given concentration of NaF (C_salt) = 0.018 M

pH = 7 + $\small \frac{1}{2}$ (3.1 + log 0.018)

or, pH = 7 + $\small \frac{1}{2}$ ( 3.1 - 1.744)

or, pH = 7 + 0.677

or, pH = 7.677

14.

a)

species ; HCN , Na ⁺ , CN^- , H⁺ ,

spectator is Na⁺ , H⁺

weak acid = HCN

CN^- is strong base .

no reaction

b.

species ; H⁺ , Br^- , K⁺ , BrO^-

HBr is strong acid and BrO^- is strong base .

spectator ; Br^- , K⁺ , H⁺​​​​​​​  

no reaction

c.

species; Na⁺ . OH^- , OCl^-

HClO is weak acid

and OH^- and OCl ^- are strong base

HClO +NaOH $\small \to$ NaOCl + H₂O

as Concentration of NaOH  is higher than that of HClO

final mixture contains excess NaOH and its salt NaOCl

15.

Buffer is a mixture of weak acid and its salt.

so only mixture of weak acid HCN and its salt NaCN will form buffer

pKa of HCN = 9.30

pH of buffer using Hasselbalch- Henderson equation is

pH = pKa + log [NaCN]/ [HCN]

OR, pH = 9.30 + log (0.10 / 0.050)

or, pH   = 9.30 + 0.3

or, pH = 9.60