This is a titration simmulation all the data is given and I\'ve listed out the g

Question

This is a titration simmulation all the data is given and I've listed out the grams for aspirin and vinegar. If you can help that would be awesome! I've worked through the rest of the problems would just like to see if my answers are correct. (compare my answers to whomever else understand titrations) thank you!

In the first section Molarity of KHP: I'm needing to find the molarity of KHP the formula is KHC8H4O4=204g the other information needed is listed. (I've already solved for the molarity and came up with 0.142 mol/L)

In the second section Standardization of NaOH: I'm needing to find the molarity of NaOH----> from my understanding to complete this section requires the molarity found in section 1.

In the third section Standardization of HCl: I need to find the molarity of HCl---> which requires the molarity found i section 2.

In the fourth section I need to find the %HC8H8O4 (ASA) in aspirin----> HC8H8O4 is 180g

In the fifth section I need to find the % acetic acid in vinegar----> HC2H3O2 (vinegar) is 60g

Using the data below, Calculate the results and average for each procedure Molarity of KHP Mass of KHP = 1.453 g Volume of solution = 50.00 ml Molarity of KHP = Standardization of NaOH trial 1 trial 2 Volume of KHP solution (ml) = 1.755 1.834 Volume of NaOH soln (ml) = 1.152 1.205 Molarity of NaOH solution average Standardization of HCl trail 1 trial 2 Volume of NaOH solution (ml) 1.488 1.578 Volume of HCl solution (ml) 1.225 1.399 Molarity of HCl average % HC_3H_3O_4(ASA) in aspirin trial 1 trial 2 Mass of tablets 1.134 g 1.128 g Volume NaOH_(aq) ml 26.05 27.02 % ASA average % acetic acid in vinegar Volume (ml) vinegar trial 1 trial 2 Volume (ml) vinegar 0.475 0.515 Volume NaOH_(aq) ml 1.915 2.115 density of vinegar = 1.005 g.ml % HC_2H_2O_2 average

Explanation / Answer

To calculate molarity of KHP, first calculate number of moles,n

n = mass taken / molecular mass = 1.453g / 204g/mol = 0.00712 moles

Molarity = Number of moles of solute / Volume of solution (in L) = 0.00712 / 0.05 = 0.142 mol / L or M

Standardisation of NaOH:

We will use the formula: M1V1 = M2V2

For trial 1, M1(KHP) = 0.142 M, V1(KHP) = 1.755 and V2(NaOH) = 1.152

M2 (NaOH) can be calculated now, 0.142 x 1.755 = M2 x 1.152

M2 = 0.216 M

Trial 2, 0.142 x 1.834 = M2 x 1.205

M2 = 0.216

Molarity of NaOh = 0.216 M

Standardisation of HCl: M2V2 = M3V3. Here M3 is molarity of HCl and M2 is molarity of NaOH calculated previously.

Trail 1, M3 = 0.216 x 1.488 / 1.225 = 0.262 M

Trial 2, M3 = 0.216 x 1.578 / 1.399 = 0.243 M

Average molarity of HCl= 0.252 M

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