A student is asked to standardize a solution of sodium hydroxide. He weighs out

Question

A student is asked to standardize a solution of sodium hydroxide. He weighs out 0.909 g potassium hydrogen phthalate (KHC9H404,treatthis as a monoprotic acid). It requires 33.6 mL of sodium hydroxide to reach the endpoint A. What is the molarity of the sodium hydroxide solution? M This sodium hydroxide solution is then used to titrate an unknown solution of perchloric acid. 8. 1121.6 mL of the sodium hydroxide solution is required to neutralize 20.4 mL of perchioric acid, what is the molarity of the perchloric acid solution? M

Explanation / Answer

Answer: First thing it is monoprotic acid and hence the reaction is 1:1 ratio .

Now the molecular mass of KHC8H4O4  = 204.2 g

So the number of moles = 0.909 /204.2 = 0.00445 mol

And Hence it is a 1:1 ratio equation menas at the end point the number of moles of monoprotic acid is equal to the number of moles of NaOH

So the number of moles of NaOH = 0.00445 mol

and the volume is = 33.6 ml

So the molarity = number of moles / volume of solution in L

= 0.00445 / 36.6 * 1000 = 0.12158 M

Hence the molarity of sodium hydroxide = 0.12158 M .

B} Now again it is a 1:1 ratio equation

number of moles 27.6 ml of NaOH is = 0.12158 * 27.6 /1000 = 0.0033556 mol

Hence at the end point the number of moles of acid and base is same

Hence the number of moles of perchloric acid is = 0.0033556 mol

and the total volume of solution is = 27.6 + 20.4 ml = 48 ml

so the molarity of perchloric acid = number od moles / volume in L

= 0.0033556 / 48 *1000 = 0.0699 M

Hence the molarity of Perchloric acid is 0.0699 M .

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