1) If 0.235g of KHP requires 16.73 mL of NaOH solution to neutralize it, calcutl

Question

1) If 0.235g of KHP requires 16.73 mL of NaOH solution to neutralize it, calcutlate the molartiry of the NaOH solution

2) If 21.32 mL of NaOH solution that was standardized in question (1) was needed to titrate 7.50 mL vinegar, calculate the weight to volume percentage of acetic acid in this sample.

3) 50.0 mL of a 1.25 M solution of nitric acid is added to 1.33 g of solid iron(III) hydroxide, what is the concentration (in M) of iron(III) ions in the final solution?

Fe(OH)3(s)  +  3 HNO3(aq)  ?  Fe(NO3)3(aq)  +  3 H2O(l)

Assume the volume of the solution remains constant.

Explanation / Answer

1)

KHP [Potassium hydrogen phthalate] has a molecular weight os 204 g

so the moles of KHP = 0.235 /204 = 0.001152.

Molarity * volume = moles.

=> M*16.73 = 0.001152.

=> M = 6.89*10^-5.

If KHP molecular weight is 71 g

then its moles = 0.235/71 = 0.0033

Molarity * volume = moles.

M*16.73 = 0.0033

M = 1.978*10^-4.

2) moles of NaOH = 1.978*10^-4*21.32 = 4.22*10^-3

as molecular mass of vinegar = 60g

weight of vinegar in solution = 60* 4.22*10^-3

= 0.253 g.

weight to volume = 0.253/7.5 *100 = 3.374 %.

3)

moles of nitric acid = 50*1.25 = 62.5*10^-3

moles of iron oxide required will be three times that of nitric acid.

so weight of iron oxide required = 0.0625/3*89 = 1.854 g

so iron oxide left =

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