Power of hydrogen ion pH-

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## Power of hydrogen ion pH-

### a) 0.01 M Ca(OH)2 or 10-2 Ca(OH)2

1 M = 2 N

[OH] =  2 x  10-2

[H+] [OH] = 10-14

[H+] = 10 -14 /[OH-]

[H+] = 10 -14 / 2 x 10 -2

[H+]  = 0.5 x 10-12  = 5 x 10 -13

pH = – log [H+]

= – log [ 5 x 10-13]

[log a x b  = log a + log b]

so,

pH  = – [log  5 + log  10-13]

[log ab = b log a]

= – [ log 5 – 13 log 10]

because , log 10 = 1

pH  =- [ 0.6990 – 13 x 1]

Therefore,

pH  = –  0.6990 + 13 x 1 ] =  12.3010

### pH = 12.3  Ans.

pH +pOH =14

pOH = 14 – pH

pOH = 14 – 12.3010 = 1.6990

## Alternative method –

[OH]  = 2 x 10-2

pOH = – log [OH]

pOH= – log [ 2 x 10-2]

[log a x b  = log a + log b]

so,

pOH  = – [log  2 + log  10-2]

[log ab = b log a]

= – [ log 2 – 2 log 10]

because , log 10 = 1

pOH  =- [ 0.3010 – 2x 1]

Therefore,

pOH  = –  0.3010 + 2x 1 ] =  1.6990

### pOH =  1.6990 Ans.

pH +pOH =14

pH = 14 – pOH

pH = 14 – 1.6990= 12.3010

## b) 0.01 N Ca(OH)2

[OH] =    10-2

[H+] [OH] = 10-14

[H+] = 10 -14 /[OH]

[H+] = 10 -14 /  10 -2

[H+]  = 10-12

pH = – log [H+]

= – log [  10-12]

[log ab = b log a]

= – [  – 12 log 10]

because , log 10 = 1

pH  =- [  – 12 x 1] = 12

### pH = 12.0  Ans.

pH +pOH =14

pOH = 14 – pH

pOH = 14 – 12.0 = 2.0

### c) 0.001 M NaOH  or 0.001 N NaOH

In both cases ,

[OH-] = 0.001 = 10-3

(because NaOH is mono acidic base. So ,  M = N)

[OH] =    10-3

[H+] [OH] = 10-14

[H+] = 10 -14 /[OH-]

[H+] = 10 -14 /  10 -3

[H+]  = 10-11

pH = – log [H+]

= – log [  10-11]

[log ab = b log a]

= – [  – 11 log 10]

because , log 10 = 1

pH  =- [  – 11 x 1] = 11

### pH = 11.0  Ans.

pH +pOH =14

pOH = 14 – pH

pOH = 14 – 11.0 = 3.0

### c)    0.002 N NaOH

[OH-] = 0.002 = 2 x 10-3

[OH] =  2 x   10-3

[OH-]  = 2 x 10-3

pOH = – log [OH]

pOH= – log [ 2 x 10-3]

[log a x b  = log a + log b]

so,

pOH  = – [log  2 + log  10-3]

[log ab = b log a]

= – [ log 2 – 3 log 10]

because , log 10 = 1

pOH  =- [ 0.3010 – 3 x 1]

Therefore,

pOH  = –  0.3010 + 3 x 1 ] =  2.6990

### pOH =  2.6990 Ans.

pH +pOH =14

pH = 14 – pOH

pH = 14 – 2.6990= 11.3010

pH =11.3 Ans.

### e) 0.006 M Mg(OH)2

It is di acidic base , So

[OH] =  2 x 0.006

= 0.012 = 1.2 x 10-2

[H+] [OH] = 10-14

[H+] = 10 -14 /[OH-]

[H+] = 10 -14 / 1.2 x  10 -2

[H+]  = 0.83  x  10-12

[H+]  = 8.3  x  10-13

pH = – log [H+]

pH= – log [ 8.3 x 10-13]

[log a x b  = log a + log b]

so,

pH  = – [log  8.3 + log  10-13]

[log ab = b log a]

= – [ log 8.3 – 13 log 10]

because , log 10 = 1

pH  =- [ 0.9191 – 13 x 1]

Therefore,

pH  = –  0.9191 + 13  =  12.08

### pH =  12.08 Ans.

pH +pOH =14

pOH = 14 – pH

pOH = 14 – 12.08 = 1.92