An analytical chemist weighs out 0.055g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. She then titrates this solution with 0.1300M NaOH solution. When the titration reaches the equivalence point, the chemist finds she has added 6.6mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

Mass of the unknown acid is 4.0g

Explanation:

The determine the molar mass of the unknown acid, the steps below can be followed

Firstly, determine the concentration of the acid, the formula below can be used;

ConcA × Va/ConcB × Vb = Na/Nb

Where ConcA is the concentration of the unknown acid

ConcB is the concentration of the NaOH base

Va is the volume of acid and Vb is the volume of base

Since, the titration was said to have reached an equivalent point, it means the number of moles of the acid (Na) was equal to the number of moles of the base (Nb) and thus both will be assumed to be 1

Thus

ConcA × 250/0.13 × 6.6 = 1/1

ConcA = 0.13 × 6.6/250

ConcA = 0.003432M

Then, determine the actual number of moles (n) of the unknown acid used,

ConcA = no of moles of acid/volume of acid (in dm³ or L)

To convert mL to L, we divide by 1000

Hence, 250ml = 0.25L

0.003432 = n/0.25

n = 0.003432 × 0.25

n = 0.01373 moles

To determine the molar mass;

n = mass/molar mass

The mass was given in the question to be 0.055g

Thus

0.01373 = 0.055/molar mass

molar mass = 0.055/0.01373

molar mass = 4.0g

i think it is true since you can test the answer and can use observations to complete it.