Question

Formic acid Ka: 1.80x10^-4

Propanoic acid Ka: 1.34x10^-5

Chloroacetic acid Ka: 1.36x10^-3

Iodic acid Ka: 1.7x10^-1

What is the pH of a solution that is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a prepared by dissolving 8.20 g of formic acid (46.03 g/mol) and 10.74 g of sodium formate (68.01 g/mol) in water and diluting to 1.00 L? pH = b 0.0800 M in propanoic acid and 0.0170 M in sodium propanoate? pH = C prepared by dissolving 3.90 g of chloroacetic acid, CICH, COOH (94.49 g/mol) in 30.0 mL of 0.5670 M NaOH and diluting to 600.0 mL? pH = d 0.0360 M in iodic acid and 0.0360 M in sodium iodate? pH =

Answer 1

Formula:-

Henderson- Hasselbalch equation for acidic buffer is

salt) pH = pke +log p" =P T y acid

ie.
(salt) pH = -log(K) +log Jacid]
....(1)

-----------------------------------

a) Answer :- 3.69

Moles of Formic acid (acid) =
8.20(9) 46.03(g/mol) = 0.178mol
  

Moles of sodium formate (salt) =
10.74(9) · = 0.158mol 68.01 (g/mol)

Now, volume of solution = 1 L

therefore,

[Formic acid] =
0.178(mol) = 0.178M 1(L)

[Sodium formate] =
0.158(mol) = 0.158M 1(L)

Kalformicacid) = 1.8 x 10-4

from eq(1)

0.158 pH = -log(1.8 x 10-4) + log100

pH = -(-3.74) + (-0.05)
.

pH = 3.74 - 0.05 = 3.69

--------------------------

b) Answer :- 4.2

[propanoic acid] (acid) = 0.08 M

[sodium propionate] (salt) = 0.017 M

K. (propanoicacid) = 1.34 x 10-5

from eq (1)

0.017 pH = -log(1.34 x 10-5)+log

pH = -(-4.87)+(-0.67)

pH = 4.87 – 0.67 = 4.2

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c) answer :- 2.71

the reaction between chloroacetic acid and NaOH is

CICHCOOH + NaOH → CICH,COONa+ H2O

we need the concentration of salt ie ClCH₂COONa which determined by ICE table

Moles of chloroacetic acid =
3.90(g) = 0.041mol 94.49(g/mol)

Moles of NaOH =
molarity (mol/L) x volume L)

=
0.567(mol/L) X 0.03(L) = 0.017(mol)

Volume of solution = 600 ml = 0.6 L

therefore,

[chloroacetic acid] =
0.041 (mol) =0.068M 0.6(L)

[NaOH] =
0.017(mol) P = 0.028M 0.6(L)

from reaction it is observed that reaction proceed by 1:1 ratio

ICE table is

	CICHCOOH	+	NaOH	$\rightarrow$	CICH,COON	+	$H_{2}O$
initial	0.068		0.028		0		0
change	0.068 - 0.028		0.028 - 0.028		+ 0.028		+ 0.028
equilibrium	0.04		0		0.028		0.028

Ka(chloroaceticacid) = 1.36 x 10-3

from eq(1)

0.028 pH = -log(1.36 x 10-3) +log 90.04

pH = -(-2.87) + (-0.16)

pH = 2.87 – 0.16 = 2.71

--------------------------

d) Answer :- 0.77  

[iodic acid] = 0.036 M

[sodium iodate] = 0.036 M

K iodicacid) = 1.7 x 10-1

frow eq(1)

0.036 pH = -log(1.7 x 10-1) +log

pH = -(-0.77) +0

pH = 0.77 + 0 = 0.77