Question

An unknown mass of sodium hydroxide is dissolved in 500 g of water. This solution is neutralized by adding successively 18 g of CH3COOH, 25.2 g HNO3, and 29.4 g of H2SO4. Determine

a) the mass of NaOH

b) the mass of H2O in the neutral solution.

Answer 1

a )

The solution contains ,

two monoprotic acids ( viz. CH3COOH = 18.0 gms and HNO3 = 25.2 gms ) & a diprotic acid H2SO4 = 29.4 gms )

Converting these weights of different acids into moles we can get the moles of H⁺ yielded by them in aqueous soutions as-

Acids......................................No. of moles.........................................................moles of H⁺ .......................................

CH3COOH ......................18 / 60 = 0.30....................................................................0.30......................................

HNO3...........................25.2 / 63 = 0.40....................................................................0.40 .......................................

H2SO4........................29.4 / 98 = 0.3 0....................................................................0.60.........................................

Thus , the total number of moles yielded by the given different acids = 1.3 moles

So, 1.3 moles of H⁺  can be neutralized by 1.3 moles of OH^- produced by ionization of 1.3 M NaOH solution.

....................................................................Na OH (aq )---------------> Na⁺ (aq ) + OH^- (aq )

........................................................................1 mole............................1 mole..................1 mole.....................

....................................................................1.3 moles......................................................1.3 moles.......................

   Now we need to have 1.3 M NaOH solution which contains NaOH = ( 40 x 1.3 )

------------------------------------------------------------------------------------- = 52.0 gms of NaOH / Litre of solution,

but actually we have used 500 g ( or = 0.50 litres ) of water, therefore the required 1.3 M NaOH solution would contain a mass of NaOH .....................................................................= 26.0 gms.

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b ) The mass of water in neutral solution can be calculated applying the relation for neutralization reactions / titrations,

................................................M1 V1 = M2 V2 ; where 1 represents acid , & 2 represents base

substituting the related values of basic solution and acidic solutions we can get the volume of water which should be used for neutralization with acidic solutions

........................................... 1.3 x V1 = 1.3 x 500

..........................................................V1 = 500 ml. [ assuming density of water = 1.00 ]

Thus the total volume of solution = Volume of NaOH solution used + Volume of acidic solutions required

.................................................... = .........................................500.00 + 500.00

....................................................= ....1000.00 ml

.............................................. or = 1.00 Litre

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