A student is given 3 beakers: Beaker 1- 50.0 ml of a solution produced by dissol

Question

A student is given 3 beakers:

Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak monoprotic acid ,HX, in enough water to produce 1 liter of solution. The empirical formula of HX is CH2O. The solution contains 3 drops of phenolphthalein.

Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

Beaker 3 – 50.0 ml of 0.250M KOH

The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.

Explanation / Answer

Step 1, -

Since 50 ml of solution in beaker1 is completly reacted with 20.0 ml of 0.250 KOH, the molarity of solution in beaker 1 ; [ using relation M1 X V1 = M2 x V2 ]

.......................................................................................................................................= ( 20.0 x 0.250 ) / 50

...................................................................................................................................... = 0.10 M

Step 2, -

Calculate Molecular mass of HX ,assuming it as = 'M'

Since, 6.0 gms per litre solution of HX is = o.1M ..............calculated as above in step 1

Therefore a solution of HX containing one gm . molecular mass ( ie. 1M ) will contain = ( 6.0 / 0.1 ) = 60 .0 gms

Step 3, -

Given Emperical formula as CH2O

emperical formula wt = 30   

n = Mol. formula wt. / Emperical formula wt

.....=60/30 = 2

So , The acid is .......( CH2O )2 or ,...CH3COOH

X ie. anion in the acid is --- CH3COO

Calculation of pH-

The pH of the solution now would depend on the molarity of acidic solution due to hydrolysis of the salt formed as a result of following reaction,

......................................... CH3COOH + KOH = CH3COOK + H2O

Using the relation M1 x V1 = M2 X V2

the molarity of the NaX ( or in this case  is CH3COOK ) solution = ( 20.0 x 0.25 ) / 70

........................................................................................................ = 0.07M

which is also the molarity of NaX solution with pH 8.8 given in beaker 2

Thus the pH of beaker 1 after addition of 20 ml = 8.8

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