A 15.00 g sample of an ionic compound NaA, where A- is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 700.0 mL of HCl was added, the pH was 5.00. The experimenter found that 1.40 L of 0.100 M HCl was required to reach the stoichiometric point of the titration

what is the molar mass of NaA? and
Calculate the pH of the solution at the stoichiometric point of the titration.

To find the molar mass of NaA, we can use the volume and concentration of HCl required to reach the stoichiometric point.

Step 1: Calculate the number of moles of HCl used at the stoichiometric point.
Volume of HCl used = 1.40 L
Concentration of HCl = 0.100 M

Number of moles of HCl = Volume × Concentration
= 1.40 L × 0.100 M
= 0.14 moles

Step 2: Find the number of moles of NaA present in the solution.
From the balanced equation, we know that one mole of HCl reacts with one mole of NaA.

Therefore, the number of moles of NaA = 0.14 moles.

Step 3: Calculate the molar mass of NaA.
Molar mass (g/mol) = Mass (g) / Moles
Mass of NaA = 15.00 g

Molar mass of NaA = 15.00 g / 0.14 moles
= 107.14 g/mol

Therefore, the molar mass of NaA is 107.14 g/mol.

To calculate the pH of the solution at the stoichiometric point, we need to consider the dissociation of the weak acid.

Let's assume that the weak acid A- is related to its conjugate base HA. We can write the dissociation equation as:

A- + H2O ⇌ HA + OH-

The reaction between A- and HCl can be written as:

A- + HCl ⇌ HA + Cl-

At the stoichiometric point, all of the A- ions are consumed. So, we have only HA and Cl- ions remaining in solution.

Step 1: Calculate the moles of HCl added at the stoichiometric point.
Volume of HCl added = 1.40 L
Concentration of HCl = 0.100 M

Moles of HCl = Volume × Concentration
= 1.40 L × 0.100 M
= 0.14 moles

Step 2: Calculate the moles of HA formed.
From the balanced equation, we know that one mole of HCl reacts with one mole of HA.

Therefore, the moles of HA = 0.14 moles.

Step 3: Calculate the concentration of HA.
Volume of solution = 100.0 mL = 0.100 L

Concentration of HA = Moles of HA / Volume of solution
= 0.14 moles / 0.100 L
= 1.40 M

Step 4: Calculate the pOH of the solution.
pOH = -log10[OH-]
OH- concentration = HA concentration

pOH = -log10(1.40) = -0.1461

Step 5: Calculate the pH of the solution.
pH = 14 - pOH
pH = 14 - (-0.1461)
pH = 14.1461

Therefore, at the stoichiometric point of the titration, the pH of the solution is approximately 14.1461.

To find the molar mass of NaA, we need to use the given information about the titration.

First, we need to determine the number of moles of HCl added to reach the stoichiometric point of the titration. We know that the volume of 0.100 M HCl required to reach the stoichiometric point is 1.40 L. So, we can calculate the number of moles of HCl using the formula:

moles of HCl = volume (in liters) × concentration (in moles per liter)

moles of HCl = 1.40 L × 0.100 mol/L = 0.140 mol

Now, let's consider the balanced equation for the reaction between NaA and HCl:

NaA + HCl → NaCl + HA

From stoichiometry, we know that 1 mole of NaA reacts with 1 mole of HCl. Therefore, the number of moles of NaA is also 0.140 mol.

Next, we can calculate the molar mass of NaA using the formula:

molar mass = mass (in grams) ÷ moles

Given that the sample mass of NaA is 15.00 g and the number of moles is 0.140 mol, we can substitute these values into the formula:

molar mass = 15.00 g ÷ 0.140 mol ≈ 107.14 g/mol

So, the molar mass of NaA is approximately 107.14 g/mol.

To calculate the pH of the solution at the stoichiometric point of the titration, we need to consider the nature of the weak acid and its conjugate base.

At the stoichiometric point, all of the NaA has reacted with HCl, leaving only the weak acid HA in solution. We can assume that the volume of the solution remains the same after adding the stoichiometric amount of HCl (1.40 L).

Since HA is the anion of a weak acid, it will undergo hydrolysis in water. This hydrolysis reaction can be represented as:

HA + H2O ↔ H3O+ + A-

At the stoichiometric point, the concentration of A- will be equal to the initial concentration of NaA (which we already calculated as 0.140 M).

To calculate the pH, we can use the equation for the hydrolysis constant, Ka:

Ka = [H3O+][A-] / [HA]

Given that the concentration of HA is equal to the initial concentration of NaA (0.140 M) and assuming that the concentration of H3O+ is negligible compared to the concentration of A-, we can simplify the equation:

Ka ≈ [A-]^2 / [HA]

Since we know the value of Ka and the concentration of A-, we can rearrange the equation to solve for [HA]:

[HA] = [A-]^2 / Ka

[HA] = (0.140 M)^2 / Ka

Now, we need to calculate the value of Ka. Since the pH at the stoichiometric point is 5.00, we can use the pH formula:

pH = -log[H3O+]

Rearranging the equation:

[H3O+] = 10^(-pH)

[H3O+] = 10^(-5.00) ≈ 1.00 × 10^(-5) M

Since [H3O+] is also equal to [A-], we can substitute this value into the equation for [HA]:

[HA] = (1.00 × 10^(-5) M)^2 / Ka

Finally, we have all the information we need to calculate the pH:

pH = -log([H3O+]) = -log([A-]) = -log([HA])

Note: The value of Ka for the weak acid A- is not provided in the given information. To calculate the pH accurately, the Ka value would need to be known.

Look at HA ionization.

HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
Solve for (H^+) = Ka(HA)/(A^-)
Note that 700 mL is exactly half way to the equivalence point; therefore, we know (HA) untitrated = (A^-) from the half titrated. Thus pKa = pH.

At th stoichiometric point you know mols HCl = M x L = 0.1M x 1.4L = 0.14mols.
That represents 15.00 g;
grams = moles/molar mass. Solve for molar mass.

15.00 g/molar mass = mols NaA and
M = moles/L at this point. I see 1.4L + 100 mL initially used.M = ?
............A^- + HOH ==> HA + OH^-
initial....?M ..............0....0
change......-x.............x......x
equil....?N-x...............x......x

Kb for A^- = (Kw/Ka for the acid) = (HA)(OH^-)/(A^-)
Substitute from the ICE chart above and solve for x = OH^-, then convert to pH.