The answer and why? 11. A solution is prepared by dissolving 9.75 g of an unknow

Question

The answer and why?

11. A solution is prepared by dissolving 9.75 g of an unknown weak base in enough water to produce 100. mL of solution. The solution is then titrated with 0.175 M HCl. The titration curve and corresponding derivatives are shown below. The molar mass (in g/mol) of the unknown weak base is closest to Hd + O 0 25 50 75 100 125 150 175 200 225 250 Volume HCI Added (ml) Volume HCI Added (ml) 50 75 100 125 150 175 200 225 250 0 25 -0.2 First Derivative Ap/(??)? -0.8 -1.0 0.10 0.05 - TYYT YTYYTTTTTTTTTTTTTTTTTTT Second Derivative ZAP/(Hd):P 0 25 50 75 100 125 150 175 200 225 250 -0.05 -0.10 | -0.15 Volume HCI Added (mL) (a) 32.8 g/mol. (b) 164 g/mol. (c) 223 g/mol. (d) 328 g/mol. (e) 656 g/mol.

Explanation / Answer

From pH Vs Volume graph (titration curve), it is clear that base is mono basic in nature. So its equivalent weight equal to molecular weight or normality equal to molarity.

From graph the range of equivalent point is 150-175 ml of 0.175 M HCl. while from first derivative its range become more narrow from 160-175 ml of HCl. Now second derivative curve it confirm that equivalence point is 170 ml to 168 ml. This point should be coincident with the inflection point (the point at which the curve changes direction).

It is tiration of weak base and strong acid.

Step I Molarity of weak base = 9.75g/Mol wt * 1000/100 = 97.5/ Molwt = MB

At equivalence point

MBVB = MAVA

VB= 100 ml, MA = 0.175 M, VA= 170 ml

Putting value in above equation.

97.5/ Mol wt * 100= 0.175* 170

Mol wt = (97.5* 100)/ (0.175*170)= 327.77 g/mol

which is close to answer (d)

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