A solution of sodiumhydroxide (NaOH) was standardized against potassium hydrogenphthalate (KHP). A known mass of KHP was titrated with
the NaOH solution until a light pink color appeared usingphenolpthalein indicator. Using the volume of NaOH requiredto neutralize KHP and the
number of moles of KHP titrated, the concentration of the NaOHsolution was calculated.
A vinegar (acetic acid) solution of unknown concentration wastitrated to the light pink endpoint with the standardized NaOHsolution. The
molarity and weight\volume % of the vinegar solution werecalculated.
A vitamin C (ascorbic acid) tablet was dissolved in approximately50 mL of distilled water and titrated with the standardized NaOHsolution. From
the results of this titration, the mg of ascorbic acid in thetablet was calculated.

Molecular formulas:

Potassium hydrogen phthalate: HKC8H4O4
Acetic acid: C2H4O2
Ascorbic acid: C6H8O6
Assume that all 3 acids are monoprotic acids.
Mass of KHP used forstandardization (g) 0.5591
Volume of NaOH required to neutralize KHP (mL) 13.39
Volume of vinegar sample titrated (mL) 5.00
Volume of NaOH required to neutralize vinegar in(mL) 8.38
Molecular weight of ascorbic acid (g/mol) 176.1271
Volume of NaOH required to neutralize ascorbic acid in Vitamin Ctablet (mL) 13.56

Calculate thefollowing

A. What is the Molecular Weight of KHP (HKC8H4O4) in g/mol?
B. How many moles of KHP were used in the standardization of theNaOH solution?
C. Calculate the concentration of NaOH solution in (mol/L).
D. Calculate the molarity of the vinegar solution (mol/L).
E. Calculate the weight/volume percentage of the vinegar solution(g/100 mL).
F. Calculate the amount of ascorbic acid in the Vitamin C tablet in(mg).

See explanation below

Explanation:

In order to solve this, we'll do it by parts, as this exercise takes some time to solve it, however, the procedure it's pretty easy to understand.

A. Molecular weight of the KHP

To do this, we need the atomic weight of each element of the KHP. These are the following:

H = 1 g/mol; K = 39 g/mol; C = 12 g/mol; O = 16 g/mol

Now, let's calculate the molecular weight. Remember to multiply the number of atoms by the atomic weight:

MM KHP = (1*5) + (39) + (4*16) + (12*8) = 204 g/mol

B. moles of KHP used

In this part, we already have the molecular weight, so, we can calculate the moles with the expression:

n = m/MM  (1)

The mass used of KHP is 0.5591 so the moles are:

n = 0.5591/204 = 2.74x10⁻³ moles

C. Concentration of NaOH

As the problem states, the KHP can be considered as a monoprotic acid, therefore, we can assume that the mole ratio between NaOH and KHP is 1:1 and we can use the following expression to calculate the concentration of the base:

MaVa = MbVb  (2)

But moles:

n = M*V  (3)

We have the moles of the KHP used, and the volume used to standarize the base, so we can solve for Molarity of the base:

Mb = na / Vb

Mb = 2.74x10⁻³ / 0.01339 = 0.2046 M

D. Molarity of vinegar solution

Using expression (2), we can calculate the vinegar solution, as we have the base volume used and volume of vinegar so:

MaVa = MbVb

Ma = MbVb/Va

Ma = 0.2046 * 8.38 / 5 = 0.3429 M

E. %W/V of vinegar

In this case, we use the following expression:

%W/V = mass solute / V solution * 100   (4)

The volume of solution would be the volume of the vinegar and volume of the base:

V solution = 8.38 + 5 = 13.38 mL

The mass of vinegar can be calculated, we have the concentration and volume, we can calculate the moles using expression (3):

n = 0.3429 * 0.005 = 1.71x10⁻³ moles

The mass of vinegar using the molecular weight of acetic acid (60 g/mol):

m = 1.71x10⁻³ * 60 = 0.1026 g

So the %:

%W/V = 0.1026/13.38 * 100 = 0.77%

F.  mg of ascorbic acid

We do the same thing as in part C and then, the mass of ascorbic acid can be calculated with the molecular weight:

MaVa = MbVb = na

na = 0.2046 * 0.01356 = 2,77x10⁻³  moles

m = 2.77x10⁻³ * 176.1271 = 0.4878 g or simply 487.8 mg

people who take vitamin supplements

explanation:

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg