Question

1)Determine the pH of each solution.

Part A

0.20 M KCHO2

Express your answer to two decimal places.

Part B

0.21 M CH3NH3I

Express your answer to two decimal places.

Part C

0.22 M KI

Express your answer to two decimal places.

2)

Part A

Rank the following compounds in order of decreasing acid strength using periodic trends.

Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.

LiH, H2O, HCl, HBr

Answer 1

Answer – 1) We are given,

Part A) [KCHO₂] = [CHO₂^-] = 0.20 M

We need to put ICE chart, since HCHO is weak acid

CHO₂^- + H₂O <------> HCOOH + OH^-

I 0.20                                0           0

C -x                                  +x         +x

E 0.20-x                           +x          +x

We know, Ka for HCOOH = 1.77*10^-4

So, Kb = 1.0*10^-14 / 1.77*10^-4

             = 5.65*10^-11

So, Kb = [HCOOH] [OH^-] / [CHO₂^-]

5.65*10^-11 = x*x / (0.20-x)

We can neglect x in 0.20-x, since Kb value is too small

So, x² = 5.65*10^-11 *0.20

x = 3.36*10^-6 M

[OH^-] =x= 3.36*10^-6 M

We know, pOH = -log [OH^-]

                         = - log 3.36*10^-6 M

                          = 5.47

pH = 14- pOH

      = 14 – 5.47

      = 8.53

Part B) We are given, [CH₃NH₃I] = [CH₃NH₃⁺] = 0.21 M

We need to put ICE chart, since CH₃NH₃⁺ is weak conjugate acid

CH₃NH₃⁺+ H₂O <------> CH₃NH₂ + H₃O⁺

I   0.21                                    0           0

C -x                                     +x         +x

E 0.21-x                               +x          +x

We know, Kb for CH₃NH₂ = 4.38*10^-4

So, Ka = 1.0*10^-14 / 4.38*10^-4

             = 2.28*10^-11

So, Kb = [CH₃NH₂] [H₃O⁺] / [CH₃NH₃⁺]

2.28*10^-11 = x*x / (0.21-x)

We can neglect x in 0.21-x, since Ka value is too small

So, x² = 2.28*10^-11 *0.21

x = 3.36*10^-6 M

[H₃O⁺] =x= 2.19*10^-6 M

So, pH = - log 2.19*10^-6 M

           = 5.66

Part C ) We are given, [KI] = 0.22 M

So, [I^-] = 0.22 M

We know HI is strong acid, so, [I^-] = [HI] = [H₃O⁺] = 0.22 M

So, , pH = - log 0.22 M

           = 0.66

2) In this one there are given the acids and we need to rank them in order of decreasing acid strength using periodic trends.

We know from the left to right in period the electronegativity increase and from the top to bottom in group it is decrease,

So Li has lowest electronegative than O, Cl and –Br , so LiH is weak acidic, since Li is easily form Li⁺ and there is H^- hydride ions released. The O is more electronegative than C- and Br, so water also weak acid than HCl and HBr.

The electronegativity for –Br is lowest than –Cl, so HBr has more acidic than HCl

So rank the given acid in the decreasing order –

HBr > HCl > H₂O > LiH