The pH for a solution is determined from the definition pH = - log[H₃O¹⁺]. A simple graphic relationship between the hydronium ion concentration and pH is shown here. The top row gives the concentration [H₃O¹⁺] (also written as H⁺ ion ) and the second line gives the matching pH . This is limited only to the whole numbered exponents of 10.

 

Table shows that at 25oC the product [H+] x [OH-] always must equal 1 x 10-14

H+		100	10-1	10-2	10-3	10-4	10-5	10-6	10-7	10-8	10-9	10-10	10-11	10-12	10-13	10-14	10-15
OH-		10-14	10-13	10-12	10-11	10-10	10-9	10-8	10-7	10-6	10-5	10-4	10-3	10-2	10-1	100	101
Kw		10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14	10-14

 

a. Hydrochloric acid, HCl(aq) is a strong acid. The [H3O1+] = 1.0 X10^-2 because strong acids are 100% ionized and the ion concentrations equal the original acid concentration. pH = -log[H3O1+]; pH = -log 1 x 10^-2 ;

The log function on a calculator gives pH = 2

or the graphical relationship shown above can be used to get the same result..

b. Nitric acid, HNO3(aq), is a strong acid, so the [H3O1+] = [ NO31-] = 1.0 x 10^-3

pH = -log[H3O1+] ; pH = -log 1 x 10^-3 ; pH = 3

 

c. Sodium hydroxide is a strong base. The Na1+ and OH1- concentrations equal the

original NaOH concentration. [Na+] = [ OH1-] = 0.1 = 1 x 10-1. Use the relationship

Kw = [H3O1+] [ OH1-] = 1 x 10-14 to calculate the [H3O1+]. Substitute 1 x 10-1 for the hydroxide concentration; 1 x 10-14 = [H3O1+] [ OH1-] = [H3O1+] [ 1 x 10-1 ] ;

[H3O1+] == 10-14--1 = 1 x 10-14+1 = 1 x 10-13 . When the coefficients are 1, a simpler method recognizes that the sum of the exponents on ten on both sides must equal -14 because [H3O1+] [ OH1-] = 1 x 10-14. This means -1 + ? = -14; The unknown exponent must be -13. so [H3O1+] = 1 x 10-13.

pH = -log[H3O1+] ; pH = -log 1 x 10-13 ; pH = 13.

d. Hydrobromic acid, HBr, ionizes 100%. The concentration of hydronium ion, [H3O1+], is 0.10 M = 1.0 x 10-1.

Using pH = -log[H3O1+]; pH = -log 1 x 10-1; pH = 1

13. d. Neutral water has equal concentrations for [H3O1+] and [ OH1-].

Kw = [H3O1+] [ OH1-] = 1 x 10-14; Let x = [H3O1+] = [ OH1-]

[ x ] [ x ] = 1 x 10-14; x2 = 1 x 10-14; x = = x 10-14÷2 = 1 x 10-7;

pH = -log[H3O1+] ; pH = -log 1 x 10-7 ; pH = 7

14. a. pH = 1 ; [H3O1+]= 10-pH; [H3O1+] = 10-1 = 1 x 10-1 or 0.1

b. pH = 0 ; [H3O1+]= 10-pH; [H3O1+] = 10-0 = 1 x 10-0 or 1

 

15. a. pH = 8 ; [H3O1+]= 10-pH; [H3O1+] = 10^-8= 1 x 10^-8;

 

Kw = [H3O1+] [ OH1-] = 1 x 10-14 ; substitute for [H3O1+].

1 x 10-8 [ OH1-] = 1 x 10-14 ;

When the multipliers are 1, a simpler method recognizes that the sum of the exponents on ten on both sides must equal -14.

This means -8 + ? = -14;

The unknown exponent must be -6. so [ OH1-] = 1 x 10-6

A more formal method is to solve for [ OH1-] .

[ OH1-] = (1 x 10-14)÷ 1 x 10-8 = (1 x 10-14)(1 x 108) = 1 x 10-14+8 = 1 x 10-6

 

[ OH1-] = 1 x 10-6 The results are the same.

 

b. pH = 10; [H3O1+]= 10-pH; [H3O1+] = 10-10= 1 x 10-10

Kw = [H3O1+] [ OH1-] = 1 x 10-14 ; [1 x 10-10 ][ OH1-] = 1 x 10-14

[ OH1-] = 1 x 10-4