So the final concentrations are F 0 6.78 times 10^-2 0.1356 2x = [OH^-] = (2 tim

Question

So the final concentrations are F 0 6.78 times 10^-2 0.1356 2x = [OH^-] = (2 times 6.78 times 10^-2) = 0.1356M p^OH = -log [0.1356] = 0.868 (watch the sig figs) pH = 14 - p^OH = 14 - 0.868 = 13.132 This is a basic pH so it makes sense. If you had the same concentration of NaOH and Ca(OH)_2 which one would have the higher pH? Why? Explain. Determine the pH of a 1.09% by mass HCl solution with a density of 1.01 g/mL. What is the mass of Hl should be present in 0.250 L of solution to obtain a solution to obtain a solution with a pH of 1.75? What is the mass of NaOH should be present in 0.250 L of solution to obtain a solution to obtain a solution with a pH of 12.36? What is the mass of Ca(OH)_2 should be present in 0.250 L of solution to obtain a solution to obtain a solution with a pH of 13.06?

Explanation / Answer

(1)

If NaOH=0.1 M

Ca(OH)2=0.1 M

For NaOH

NaOH----->Na++OH-

So [NaOH]=[OH-]=0.1

So pOH=-log[OH-]=-log 0.1=1

We know that

pH+pOH=14

So pH=14-1=13

Now for Ca(OH)2

Ca(OH)2------------->Ca2++2OH-

So

[Ca(OH)2]=0.1 M

[OH-]=2*0.1=0.2 M

pOH=-log[OH-]=log0.2=0.698

So

pH=14-0.698=13.302

So pH of Ca(OH)2 is higher than NaOH

(2)

1.0.015 M HNO2

HNO2+H2O------->H3O++NO3

Initial 0.015 0

change -0.015 0.015

Final 0 0.015

So

[H3O+]=[H+]=0.015 M

So pH=-log[H+]=-log0.015=1.824

pOH=14-1.823=12.176

[OH-] =10-12.716=6.67*10-13 M

2.0.0052 M HBr

HBr+H2O-------->H3O++Br-

So [HBr]=[H+]=[H3O+]=0.0052 M

So

pH=-log0.0052=2.284

pOH=14-2.284=11.716 M

[OH-]=10-11.716=1.92*10-12 M

3 .0.087 M HI and 0.125 M HBrO4

HI is strong acid compare to HBrO4

so we consider HI

[H+]=0.087 M

pH=1.060

pOH=12.94

[OH-]=1.14*10-13

4. 0.0112 M Ba(OH)2

Ba(OH)2------------>Ba2++2OH-

So [OH-]=2*0.0112=0.0224 M

So pOH=-log0.0224=1.649

pH=12.35

[H3O+]=4.46*10-13

5. 4.48*10-4 M NaOH

[OH-]=4.48*10-4 M

pOH=3.3487

pH=10.651

[H3O+]=2.23*10-11

6.

1.5*10-3 M Ca(OH)2

[OH-]=3*10-3 M

pOH=2.5228

pH=11.477

[H3O+]=3.33*10-12

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