A 5.75-g quantity of a diprotic acid was dissolvedin water andmade up to exactly

Question

A 5.75-g quantity of a diprotic acid was dissolvedin water andmade up to exactly 275 mL. Calculate the molar mass of the acid if25.0 mL of the solution required 11.0 mL of 1.00 M KOH forneutralizaion. Number=_____________g/mol ********WILL RATE********* A 5.75-g quantity of a diprotic acid was dissolvedin water andmade up to exactly 275 mL. Calculate the molar mass of the acid if25.0 mL of the solution required 11.0 mL of 1.00 M KOH forneutralizaion. Number=_____________g/mol ********WILL RATE*********

Explanation / Answer

You can start out by noting that in a titration, moles base = molesacid. MHVH = MOHVOH (MH)25.0mL = 1.00M * 11.0mL MH = 0.440mol H+ Keep in mind that this is diprotic, so while you have 0.440molhydrogen ion, you actually only have 0.220mol acid. This 0.220mols of acid in 25mL is the same concentration as 5.75gin 275mL. Note that grams divided by molar mass is units ofmoles. 0.220mol / 25.0mL = (5.75g / Mm) / 275mL Mm = (5.75g / 275mL) / (0.220mol / 25.0mL) =2.38g/mol Check the math and also check the numbers you're given. While I'mcertain this is the correct way to do it, 2.38 g/mol just doesn'tmake sense. Let me know what you think, and I'd be happy to take asecond look =)

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