An aqueous solution is 35 by mass methylamine CH3NH2 and ha

An aqueous solution is 35 % by mass methylamine (CH3NH2) and has a density of 0.85 g/cm3. What is the pH of the resulting solution if 50 mL of this solution is diluted to 1000 mL Kb = 0.00036 .

Solution

An aqueous solution 35 % by mass means 35 gr in 100 ml solution. From this we can calculate molarity of the solution. M = (wt/m.wt )1000/V(ml) = (35/31)1000/100 = 11.29 M Now, 50 ml of this solution is diluted to 1000 ml, i.e. M1V1=M2V2 M2 = M1V1/V2 = 11.29* 50/1000 = 0.56M given base is weak base so, weak base pOH = 1/2(pKb+log C)

by substituting the values in above equation ,

pOH = 4.18

pH=14-pOH

=14-4.18

= 9.82