1.

Calculate the PH of the resultant mixture: 10 Ml of 0.1M H2SO4 + 10 Ml 0.1M KOH

Answer»

Number of moles of H+ ion = 2 x number of moles of H2SO4 

 =2 x M(V/1000) 

= 2 x 0.1x10/1000 

= 2x 10-3 

Number of moles of OH ion = number of moles of NaOH 

 = M(V/1000) 

 = 0.1x10/1000 

=1 x 10-3 

Number of moles of H+ ion after neutralization=2 x 10-3 - 1x 10-3 

 = 1x 10-3 

Molarity of H+ ion= n x1000/V 

 = 1x 10-3 x1000/20 

 =0.05 

 PH = - log(H+ ) 

 = -log (0.05) 

 = 1.301 


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