Power of hydrogen ion

source :designerwater.co.za

## Power of hydrogen ion ‘pH’-

### Question 1) Calculate pH(Power of hydrogen ion )of

### a)10^{-3} N HNO_{3} ( b) 10^{-3} M H_{2}SO_{4} . (c) 10^{-3} N H_{2}SO_{4} (d) 0.01 N HCl (e) 10 ^{-8} N HCl (f) 10^{-8} N NaOH

### Solution )

### Strong acid ionises completely at normal dilution.

## a) 10^{-3} HNO_{3}

*[H ^{+}] = 10^{-3}*

*pH = – log [H ^{+}]*

*= – log [ 10 ^{-3}]*

*pH = – [- 3 log 10 ]*

*because , log 10 = 1*

*Therefore,*

*pH = – [- 3 x 1 ] = 3*

*pH = 3 Ans.*

## b) 10^{-3} M H_{2}SO_{4}

*H _{2}SO_{4} is dibasic acid,*

*1 M = 2N*

*[H ^{+}] = 2 x 10^{-3}*

*pH = – log [H ^{+}]*

*= – log [ 2 x 10 ^{-3}]*

*[log a x b = log a + log b]*

*so,*

*pH = – [log 2 + log 10 ^{-3}]*

*[log a ^{b} = b log a]*

*= – [ 0.3010 – 3 log 10]*

*because , log 10 = 1*

*p H =- [ 0.3010 – 3 x 1]*

*Therefore,*

*pH = – 0.3010 + 3 x 1 ] = 2.6990*

*pH = 2.6990 Ans.*

## c)10^{-3} NH_{2}SO_{4}

*Normality of strong acid = [H ^{+}]*

*[H ^{+}] = 10^{-3}*

*pH = – log [H ^{+}]*

*= – log [ 10 ^{-3}]*

*pH = – [log 10 ^{-3}]*

*= – [ – 3 log 10]*

*because , log 10 = 1*

*p H =- [- 3 x 1] = 3*

*Therefore,*

*pH (Power of hydrogen ion ) = 3.0 Ans.*

## d) 0.01 N HCl

*[H ^{+}] = 0.01 = 10^{-2}*

*pH = – log [H ^{+}]*

*= – log [ 10 ^{-2}]*

*pH = – [- 2 log 10 ]*

*because , log 10 = 1*

*Therefore,*

*pH = – [- 2 x 1 ] = 3*

### pH = 2 Ans.

## e) 10^{-8} N HCl

*[H ^{+}] = 10^{-8}*

*pH = – log [H ^{+}]*

*= – log [ 10 ^{-8}]*

*pH = – [- 8 log 10 ]*

*because , log 10 = 1*

*Therefore,*

*pH = – [- 8 x 1 ] = 8*

*pH (Power of hydrogen ion )= 8*

*But pH =8 is not possible in case of HCl because it is an acid and pH of acid should be less than 7.*

*Correct solution –*

*[H ^{+}] [OH^{–}] = 10^{-14}*

*In case of water,*

*[H ^{+}] = [OH^{–}]*

*[H ^{+}] of water = 10^{-7} ( not neglected )*

*So,*

*Total [H ^{+}] present in 10^{-8} N HCl = 10^{-8} + 10^{-7}*

*[H ^{+}] = 0.1 x 10^{-7} + 1 x 10^{-7}*

*[H ^{+}] =1.1 x 10 ^{-7}*

*pH = – log [H+]*

*= – log [ 1.1 x 10 ^{-7}]*

*[log a x b = log a + log b]*

*so,*

*pH = – [log 1.1 + log 10 ^{-7}]*

*[log a ^{b} = b log a]*

*= – [ 0.04139 – 7 log 10]*

*because , log 10 = 1*

*p H =- [ 0.04139 – 7 x 1]*

*Therefore,*

*pH = – 0.04139 + 7 x 1 ] = 6.9586*

*pH = 6.9586 Ans.*

## f) 10^{-8} NaOH

*[OH ^{–}] = 10^{-8}*

*[H ^{+}] [OH^{–}] = 10^{-14}*

*[H ^{+}] = 10 ^{-14} /[OH-]*

*[H ^{+}] = 10 ^{-14} / 10 ^{-8}*

*[H ^{+}] = 10 ^{-6}*

*pH = – log [H ^{+}]*

*= – log [ 10 ^{-6}]*

*pH = – [- 6 log 10 ]*

*because , log 10 = 1*

*Therefore,*

*pH = – [- 6 x 1 ] = 6*

*pH = 6*

*But pH =6 is not possible in case of NaOH because it is a base and pH(Power of hydrogen ion )of base should be more than 7.*

*Correct solution –*

*[H ^{+}] [OH^{–}] = 10 -14*

*In case of water,*

*[H ^{+}] = [OH^{+}]*

*[OH ^{–}] of water = 10 ^{-7} ( not neglected )*

*So,*

*Total [OH ^{–}] present in 10^{-8} N NaOH = 10^{-8} + 10^{-7}*

*[OH ^{–}] = 0.1 x 10^{-7} + 1 x 10^{-7}*

*[OH ^{–}] =1.1 x 10 ^{-7}*

*pOH = – log [OH ^{–}]*

*= – log [ 1.1 x 10 ^{-7}]*

*[log a x b = log a + log b]*

*so,*

*pOH = – [log 1.1 + log 10 ^{-7}]*

*[log a ^{b} = b log a]*

*= – [ 0.04139 – 7 log 10]*

*because , log 10 = 1*

*pOH =- [ 0.04139 – 7 x 1]*

*Therefore,*

*pOH = – 0.04139 + 7 x 1 ] = 6.9586*

*pH + pOH = 14*

*pH = 14 -pOH*

*= 14 – 6.9586*

*pH = 7.0414 Ans.*