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miskamm [114]
3 years ago
8

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

ith distilled water. The technician then titrates this weak acid solution with 0.0969 M KOH . She reaches the endpoint after adding 43.81 mL of the KOH solution. Determine the number of moles of the weak acid in the solution.
Chemistry
1 answer:
Elis [28]3 years ago
6 0

<u>Answer:</u> The number of moles of weak acid is 4.24\times 10^{-3} moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}

We are given:

Volume of solution = 43.81 mL = 0.04381 L      (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol

The chemical reaction of weak monoprotic acid and KOH follows the equation:

HA+KOH\rightarrow KA+H_2O

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, 4.24\times 10^{-3}mol of KOH will react with = \frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol of weak monoprotic acid.

Hence, the number of moles of weak acid is 4.24\times 10^{-3} moles.

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3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.
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Volume percent = Volume of solute
                              ----------------------------------
                                Volume of the solution
  
                  2                    Volume of the solute
               -------   =           ------------------------------
                100                               250
                     
         Volume of the solute = 2 x 250
                                                ------------
                                                  100         

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Hope this helps!




                            




4 0
2 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
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NaOH + CH3COOH → CH3COONa + H2O 

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pH = 4.60.

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