A student finds that 34.6 mL of 0.104 M HCI is required to titrate 20.2 mL of a

Question

A student finds that 34.6 mL of 0.104 M HCI is required to titrate 20.2 mL of a NaoH solution to a phenolphthalein endpoint. What is the molarity of NaOH the solution? To standardize a solution of NaOH, a student measured 0.326 g of potassium hydrogen phthalate (KHC_8 H_4 O_4 or KHP) and dissolved it in 50.0 mL of water. To the solution, two drops of phenolphthalein indicator was added. To reach the endpoint of the titration where the solution retained a pale pink color, a volume of 15.3 mL was required. What is the molarity ofthe NaOH solution? (KHP is a monoprotic acid that furnishes one H^+ ion; at the endpoint mols base = mol acid) A sample containing 0.450 g of nicotinic acid, a monoprotic acid, required 34.74 mL of 0.1053 M NaOH to titrate the sample to phenolphthalein endpoint. What is the molar mass of the nicotinic acid?

Explanation / Answer

1. Molarity of NaOH is 0.178 M

Given, M1 = 0.104 M, V1 = 34.6 mL

           M2 = ?, V2 = 20.2 mL

We know M1V1 = M2V2

Or, 0.104 M x 34.6 mL = ? x 20.2 mL

3.598 M mL/ 20.2 mL = ?

0.178 M = ?

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