200 mL of decimolar H2SO4 is mixed with 200 ml of semimolar NaOH, find the total MV.
Asked by tejas pandey | 4th Aug, 2012, 10:16: AM
Expert Answer:
The formula use will be: MH2SO4VH2SO4 + MNaOHVNaOH = MfinalVfinal
The no. of moles of 0.1 M sulphuric acid or MH2SO4VH2SO4 = 0.1 M x 0.2 l = 0.02 moles
The no. of moles of 0.5 M NaOH or MNaOHVNaOH = 0.5 M x 0.2 l = 0.1 moles
So, the total no. of moles or one can say MfinalVfinal = 0.02 + 0.1 = 0.12 moles
The final volume = 0.2 litres + 0.2 litres = 0.4 litres
So, the molarity of mixture obtained after mixing = total no. of moles/total volume
= 0.12/0.4
= 0.3 M
The formula use will be: MH2SO4VH2SO4 + MNaOHVNaOH = MfinalVfinal
The no. of moles of 0.1 M sulphuric acid or MH2SO4VH2SO4 = 0.1 M x 0.2 l = 0.02 moles
The no. of moles of 0.5 M NaOH or MNaOHVNaOH = 0.5 M x 0.2 l = 0.1 moles
So, the total no. of moles or one can say MfinalVfinal = 0.02 + 0.1 = 0.12 moles
The final volume = 0.2 litres + 0.2 litres = 0.4 litres
So, the molarity of mixture obtained after mixing = total no. of moles/total volume
= 0.12/0.4
= 0.3 M
Answered by | 30th Aug, 2012, 12:54: PM
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