Question

A student prepares a weak acid solution by dissolving 0.3000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.1025 M NaOH. Calculate the molar mass of the acid. g/mol (a) Would the molar mass be too high, too low, or unaffected if the student accidentally used 0.4000 g in the calculation? Explain. since the molar mass and mass of acid are The molar mass would be -elect select- ▼ (b) Would the molar mass be too high, too low, or unaffected if the student accidentally used 23.5 mL instead of 33.5 mL? Explain. The molar mass would be -Select -since the molar mass and number of moles which depends on the volume are

Answer 1

The molar mass of weak acid HZ is calculated as 87.3 g/mol based on the titration with NaOH. If the mass of HZ used is incorrect, the calculated molar mass would be inaccurate; too high if more mass is used, and too low if a smaller titration volume is used.

Step-by-step explanation:

To calculate the molar mass of the acid HZ, we start by finding the number of moles of NaOH used in the titration. The number of moles of NaOH is calculated as:

Number of moles of NaOH = 0.0335 L × 0.1025 mol/L = 0.0034375 mol

Because HZ is a weak acid, we can assume that it reacts with NaOH in a 1:1 molar ratio:

1 mol HZ + 1 mol NaOH → Salt + H2O

Hence, the moles of HZ will also be 0.0034375 mol. To find the molar mass of HZ, we use the initial mass of the acid HZ:

Molar Mass of HZ = Mass of HZ / Number of moles of HZ
= 0.3000 g / 0.0034375 mol
= 87.3 g/mol

(a) If the student used 0.4000 g instead of 0.3000 g, the calculated molar mass would be too high. Since molar mass is calculated using the mass of the sample, increasing the mass without changing the number of moles of acid would result in a higher calculated molar mass.

(b) If the student used 23.5 mL instead of 33.5 mL, the calculated molar mass would be too low. Since fewer moles of NaOH would be used, it would seem that less HZ was neutralized, implying that the molar mass of HZ is smaller than it actually is.

Answer 2

87.37g/mol

A. 116.49g/mol

The molar mass of HZ increased

B. 124.5459 g/mol

The molar mass of HZ increased.

Step-by-step explanation:

For a neutralisation reaction,

HZ(aq) + NaOH(aq) --> NaZ + H2O(l)

So from the above equation, 1 mole of acid, HZ reacted with 1 mole of base, NaOH to give 1 mole of salt.

Calculating moles of NaOH = molarity(molar concentration) * volume of solution

Molarity is defined as the number of moles of solute per liter of solution.

Moles of NaOH = 33.5 x 10^-3 * 0.1025

= 3.43375 x 10^-3 mol

Using stoichiometry, therefore there were 3.43375 x 10^-3 mol of HZ reacted.

Number of moles = mass/molar mass

Therefore, molar mass = 0.300/3.43375 x 10^-3 mol

= 87.37g/mol

Molar mass of a chemical compound is defined as the mass of a sample of that compound per the amount of substance in that sample in moles.

A. Molar mass = 0.400/3.43375 x 10^-3

= 116.49g/mol

The molar mass of HZ increased.

B. Moles of NaOH = 23.5 x 10^-3 * 0.1025

= 2.40875 x 10^-3 mol

By stoichiometry, therefore 2.40875 x 10^-3 mol of HZ reacted.

Number of moles = mass/molar mass

Therefore, molar mass = 0.300/2.40875 x 10^-3

= 124.5459 g/mol

The molar mass of HZ increased.