A chemistry student weighs out 0.154 g of chloroacetic acid (HCH2CICO2) into a 250. mL volumetric flask

5 answers
Question:

A chemistry student weighs out 0.154 g of chloroacetic acid (HCH2CICO2) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1400 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.

Answers

11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.

Explanation:

Chloroacetic acid is an monoprotic acid.

Neutralization reaction: [tex]ClCH_{2}COOH+NaOH\rightleftharpoons ClCH_{2}COONa+H_{2}O[/tex]

So, 1 mol of chloroacetic acid is neutralized by 1 mol of NaOH.

Molar mass of chloroacetic acid = 94.5 g/mol

So, 0.154 g of chloroacetic acid = [tex]\frac{0.154}{94.5}[/tex] moles of chloroacetic acid

                                                     = 0.00163 moles of chloroacetic acid

Lets assume V mL of 0.1400 M of NaOH is required to reach equivalence point.

So, number of moles of NaOH needed to reach equivalence point

      = [tex]\frac{0.1400\times V}{1000}[/tex] moles

So, [tex]\frac{0.1400\times V}{1000}=0.00163[/tex]

or, V = 11.6

Hence, 11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.

28 mL of NaOH

Explanation:

First, we need to write the overall acid base reaction:

HC₃H₅O₃ + NaOH <> NaC₃H₅O₃ + H₂O

According to the above reaction, we can see that this reaction is already balanced and that the mole ratio between the acid and the base is 1:1. Therefore we can say the following:

n₁ = n₂  (1)

1 would be the acid, and 2 the base.

Now, in an acid base titration, when the titration reach the equivalence point, in this point, moles of each reactant is the same, and we use expression (1), but we can also use concentration and volume of both reactants. so, if we want to know the volume of the base, we need to rearrange the  above expression into an expression that use concentration and volume, which is the following:

n = M * V  (2)

If we replace this expression into (1) we have:

M₁V₁ = M₂V₂  (3)

From there, we just solve for the volume.

V₂ = M₁V₁ / M₂    or  V₂ = n₁ / M₂  (4)

And this is how we can calculate the volume of the base. Now, we need to know the moles used of the acid. For that, we need the molecular weight of the lactic acid which is 90.08 g/mol, so the moles are:

n₁ = 0.126 / 90.08 = 1.4x10⁻³  moles

Now, we have these moles, we can use (4) to get the volume of the base:

V₂ = 1.4x10⁻³ / 0.05

V₂ = 0.028 L or 28 mL of NaOH

7.65 × 10⁻³ L.

3 sig. fig. as in the mass of formic acid.

Explanation

How many moles of formic acid?

Relative atomic mass from a modern periodic table:

H- 1.008,C- 12.011, andO- 15.999.

Molar mass of formic acid [tex]\text{HCOOH}[/tex]:

[tex]1.008 + 12.011 + 2\times 15.999 + 1.008 = 46.025\;\text{g}\cdot \text{mol}^{-1}[/tex].

Number of moles of formic acid molecules in the 0.0634 gram sample:

[tex]\displaystyle n = \frac{m}{M} = \frac{0.0634}{46.025} = 1.37751\times 10^{-3}\;\text{mol}[/tex].

Keep at least one extra sig. fig. to avoid rounding errors.

How many moles of NaOH?

It takes one [tex]\text{OH}^{-}[/tex] ion to neutralize each carbonyl group [tex]-\text{COOH}[/tex].

There's one carbonyl group in each formic acid [tex]\text{H-}\text{COOH}[/tex] molecule. Each formula unit of NaOH supplies one [tex]\text{OH}^{-}[/tex] ion. As a result, it takes one mole formula units of NaOH to neutralize one mole formic acid molecules.

There are [tex]1.37751\times 10^{-3}\;\text{mol}[/tex] of formic acid in the volumetric flask. It will take the same number of NaOH formula units to reach the equivalence point.

How many liters of NaOH?

[tex]\displaystyle V = \frac{n}{c} = \frac{1.37751\times 10^{-3}\;\text{mol}}{0.1800\;\text{M}}= \frac{1.37751\times 10^{-3}\;\text{mol}}{0.1800\;\text{mol}\cdot\text{L}^{-1}} = 7.65\times 10^{-3}\;\text{L}[/tex].

How many significant figures?

Number of sig. fig. in data involved in the calculation:

3 in the mass of the formic acid,4 in the concentration of NaOH.

Take the least number of sig. fig. as the number of sig. fig. in the answer. As a result, the volume of NaOH should be rounded to three sig. fig.

THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L

Explanation:

The titration is between citric acid (H3C6H507) and NaOH

mass of citric acid = 0.306 g

Volume of citric acid = 250 mL = 250 /1000 = 0.25 L

Concentration of NaOH = 0.1000 M

Volume = unknown

First calculate the molar mass of citric acid

( 1 * 3 + 12* 6 + 1*5 + 16*7) = (4 + 72 + 5 + 112) = 193 g/mol

Since,

Concentration in moles/dm3 = concentration in g/dm3 / RMM

So the molarity of citric acid is:

Molarity = 0.306g / 0.25dm3  / Rmm

Molarity = 1.224g/dm3 / 193 g/mol

Molarity = 0.0063 M

Equation for the reaction is:

C3H5O(COOH)3 + 3NaOH → Na3C3H5O(COO)3 + 3H2O

Using the formula:

CaVa / CbVb = na/ nb

Ca = 0.0063 M

Cb = 0.1000 M

Va = 0.25 L

Vb = unknown

na = 1

nb = 3

Vb = Ca Va nb/ Cb na

Vb = 0.0063 * 0.25 * 3 / 0.1000 * 1

Vb = 0.4725 / 0.1000

Vb = 4.725 L

The volume of NaOH needed to reach the equivalent point is therefore 4.725 L

the volume of NaOH solution the student will need to add to reach the equivalence point is 0.028 L

Explanation:

the solution is in the attached Word file

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