kramtus51

2022-01-04

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutra lize a 0.2053-g sample of a weak acid. What is the molar mass of the acid if it is monoprotic?

Ben Owens

Molarity of $NaOH=0.1008M=0.1008mo\frac{l}{L}$
Volume of solution $=15mL=0.015L\left(1mL=0.001L\right)$
Using the molarity and volume of the solution, we can determine the moles of NaCl.

1 mol of NaOH will neutralize 1 mol of ${H}^{+}$ in a monoprotic acid. So 0.001512mol NaOH will neutralize 0.001512mol ${H}^{+}$
1 mol of monoprotic acid contains 1 mol ${H}^{+}$. So 0.001512mol ${H}^{+}$ is released by 0.001512mol monoprotic acid.
Moles of monoprotic acid $=0.001512mol$
Mass of the monoprotic acid $=0.2053g$
Using this equation, we can determine the molar mass.

$=\frac{0.2053g}{0.001512mol}$
$=135.7\frac{g}{m}ol$

Fasaniu

Mol of NaOH $=15.0mL×\left(\frac{1L}{1000mL}\right)×\left(\frac{0.1008mol}{1L}\right)=1.51×{10}^{-3}mol$

xleb123

Given information:
Volume of sodium hydroxide (${V}_{\text{NaOH}}$) = 15.0 mL
Concentration of sodium hydroxide (${C}_{\text{NaOH}}$) = 0.1008 M
Mass of weak acid (${m}_{\text{acid}}$) = 0.2053 g
We can start by calculating the number of moles of sodium hydroxide used in the neutralization reaction.
${n}_{\text{NaOH}}={C}_{\text{NaOH}}×{V}_{\text{NaOH}}$
Substituting the given values:
${n}_{\text{NaOH}}=\left(0.1008\phantom{\rule{0.167em}{0ex}}\text{M}\right)×\left(15.0\phantom{\rule{0.167em}{0ex}}\text{mL}\right)$
Next, we need to determine the number of moles of the weak acid. Since the weak acid is monoprotic, it reacts with one mole of sodium hydroxide (NaOH) in the neutralization reaction.
${n}_{\text{acid}}={n}_{\text{NaOH}}$
Now we can calculate the molar mass of the acid using the formula:

Substituting the given values:

Now, let's substitute the value of ${n}_{\text{NaOH}}$ and calculate the molar mass of the acid.

Simplifying this expression gives us the molar mass of the acid.

fudzisako

To solve the problem, we can use the concept of stoichiometry and the equation for neutralization reactions.
The balanced chemical equation for the neutralization reaction between sodium hydroxide ($\text{NaOH}$) and a monoprotic weak acid ($\text{HA}$) can be written as:
$\text{HA}+\text{NaOH}\to \text{NaA}+{\text{H}}_{2}\text{O}$
From the given information, we know that 15.0 mL of 0.1008 M sodium hydroxide solution is needed to neutralize a 0.2053-g sample of the weak acid.
We can start by calculating the number of moles of sodium hydroxide used in the reaction:

Next, we need to determine the molar ratio between sodium hydroxide and the weak acid. Since the weak acid is monoprotic, the molar ratio is 1:1. This means that one mole of sodium hydroxide reacts with one mole of the weak acid.
Now, we can calculate the number of moles of the weak acid:

Given that the molar mass ($M$) is defined as the mass (in grams) divided by the number of moles, we can rearrange the equation to solve for the molar mass of the weak acid:

Plugging in the values, we have:

Substituting the moles of NaOH calculated earlier, we get:
$M=\frac{0.2053\phantom{\rule{0.167em}{0ex}}\text{g}}{0.1008\phantom{\rule{0.167em}{0ex}}\text{M}×0.0150\phantom{\rule{0.167em}{0ex}}\text{L}}$

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