Question

question 1
suppose that 1.25 g of ba(hc8h4o4)2 (barium hydrogen
phthalate) is titrated to the equivalence point by 0.0150 l of a
naoh solution. what is the molarity of the naoh solution?
the equation is: ba(hc8h4o4)2 + 2naoh → na2c8h4o4 + h2o
question 2
what is the molar mass of an unknown acid, assume x is not small. if 0.367 g of the
acid is used to neutralize 0.0423 l of a 0.145 m fe(oh)2 solution?
the equation is: 2h3c + 3fe(oh)2 → fe3c2 + 6h2o
question 3
find the ph of a solution that contains 4.35 g of h2se dissolved in
3.15 liters of solution. this is considered a strong acid.
the equation is: h2se → 2h+ + se2-
question 4
calculate the poh of 0.40 m h2so3 (ka = 1.5 x 10-7)
during your calculations, assume x is small.
the equation is: h2so3 + h2o ⇌ hso3-1 + h3o+1
question 5
what mass of hno3 (ka = 3.1 x 10-5) will produce 3.00 l of a
solution having a ph of 4.75? the acids molar mass is
63.01 g/mol.
during your calculations, assume x is small.
the equation is: hno3 + h2o ⇌ no3-1 + h3o+1
question 6
calculate the ph of 0.512 m c6h5nh2 (ka = 2.3 x 10-5)
during your calculations, assume x is small.
the equation is: c6h5nh2 + h2o ⇌ c6h5nh3+1 + oh-1
question 7
calculate the initial concentration of a solution of c5h5n
(kb = 5.9 x 10-6) which has a ph of 11.40.
during your calculations, assume x is small.
the equation is: c5h5n + h2o ⇌ c5h5nh+1 + oh-1
question 8
what is the ph of a 0.60 m solution of hno2 (ka = 4.5 x 10-4)?
during your calculations, assume x is not small.
the equation is: hno2 + h2o ⇌ no2-1 + h3o+1

Explanation:

Let's solve Question 1 step by step:

Step 1: Determine the moles of \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \)

First, we need to find the molar mass of \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \). The molar masses are: \( \text{Ba} = 137.33 \, \text{g/mol} \), \( \text{H} = 1.008 \, \text{g/mol} \), \( \text{C} = 12.01 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).

For \( \text{HC}_8\text{H}_4\text{O}_4 \):

• \( \text{H} \): \( 1 \times 1.008 = 1.008 \)
• \( \text{C} \): \( 8 \times 12.01 = 96.08 \)
• \( \text{H} \): \( 4 \times 1.008 = 4.032 \)
• \( \text{O} \): \( 4 \times 16.00 = 64.00 \)
• Total for \( \text{HC}_8\text{H}_4\text{O}_4 \): \( 1.008 + 96.08 + 4.032 + 64.00 = 165.12 \, \text{g/mol} \)

For \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \):

• \( \text{Ba} \): \( 137.33 \)
• \( 2 \times \text{HC}_8\text{H}_4\text{O}_4 \): \( 2 \times 165.12 = 330.24 \)
• Total molar mass: \( 137.33 + 330.24 = 467.57 \, \text{g/mol} \)

Now, moles of \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \) = \( \frac{\text{mass}}{\text{molar mass}} = \frac{1.25 \, \text{g}}{467.57 \, \text{g/mol}} \approx 0.002673 \, \text{mol} \)

Step 2: Use the stoichiometry from the balanced equation

The balanced equation is: \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 + 2\text{NaOH}
ightarrow \text{Na}_2\text{C}_8\text{H}_4\text{O}_4 + \text{H}_2\text{O} \) (Wait, actually, the correct balanced equation should be: \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 + 2\text{NaOH}
ightarrow \text{Ba(C}_8\text{H}_4\text{O}_4\text{)}_2 + 2\text{H}_2\text{O} \)? Wait, no, the given equation is \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 + 2\text{NaOH}
ightarrow \text{Na}_2\text{C}_8\text{H}_4\text{O}_4 + \text{H}_2\text{O} \)? Wait, maybe a typo, but the stoichiometric ratio is 1:2 between \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \) and \( \text{NaOH} \).

So moles of \( \text{NaOH} \) = \( 2 \times \) moles of \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \) = \( 2 \times 0.002673 \, \text{mol} = 0.005346 \, \text{mol} \)

Step 3: Calculate the molarity of \( \text{NaOH} \)

Molarity \( M = \frac{\text{moles of solute}}{\text{volume of solution (in L)}} \)

Volume of \( \text{NaOH} \) solution is \( 0.0150 \, \text{L} \)

So \( M = \frac{0.005346 \, \text{mol}}{0.0150 \, \text{L}} \approx 0.3564 \, \text{M} \)

Wait, let's check the molar mass again. Wait, maybe I made a mistake in the formula. Let's re-calculate the molar mass of \( \text{Ba(HC}_8\text{H}_4\text{O}_4\text{)}_2 \):

\( \text{Ba} \): 137.33

\( \text{HC}_8\text{H}_4\text{O}_4 \): H (1) + C8 (8×12.01=96.08) + H4 (4×1.008=4.032) + O4 (4×16=64) = 1 + 96.08 + 4.032 + 64 = 165.112

So two of them: 2×165.112 = 330.224

Plus Ba: 137.33 + 330.224 = 467.554 g/mol. So 1.25 g / 467.554 g/mol ≈ 0.002673 mol. Then moles of NaOH: 2×0.002673 = 0.005346 mol. Volume is 0.0150 L. So molarity is 0.005346 / 0.0150 ≈ 0.356 M.

Answer:

The molarity of the \( \text{NaOH} \) solution is approximately \( \boldsymbol{0.356 \, \text{M}} \) (or more precisely, let's do the calculation again:

1.25 g / 467.55 g/mol = 0.002673 mol (Ba(HC8H4O4)2)

Moles of NaOH: 2 * 0.002673 = 0.005346 mol

Molarity: 0.005346 mol / 0.0150 L = 0.3564 M ≈ 0.356 M)