Question

A student titrated a 1.0115 g sample of potassium hydrogen phthalate (hkc8h4o4 204.2285g/mol with sodium hydroxide. the initial buret reading was 1.16 ml and final reading 40.62 ml write a balanced equation

Answer 1

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first determine the equation. KHP is a monoprotic acid, so it reacts with NaOH in a 1:1 ratio. 

a) HKC8H4O4 + NaOH → NaKC8H4O4 + H2O 

This allows you to calculate the molarity of NaOH, from the equation V1M1n2 = V2M2n1 

Molarity times volume will give you the number of moles of acid. You can work this out by dividing the mass used by the molar mass. For this question, you will have 1.0115/204.2285 moles KHP, so this value can be substituted for V1M1 in the equation. You need 38.46 mL of NaOH to titrate all the acid, so this will be V2. Now you find M2. 

b) 1.0115/204.2285 × 1 = 38.46 × M2 × 1, so M2 = 0.0001288M 

Now go on to part 3. Here, you use a similar calculation. The acid and base react in a 1:1 ratio, so n1 = n2 = 1. M2 = 0.1905 and V2 = 25.62. V1M1 will be calculated. 

V1M1 × 1 = 25.62×0.0001288×1 = 0.003299856 

Now mass divided by molar mass = V1M1, so molar mass = mass/V1M1. 

1.2587/0.003299856 = 381.44 g/mol