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Ray Of Light [21]
3 years ago
10

In a titration experiment 12.5 ml of 0.500 m h2so4 neutralized 50.0 ml of naoh. the concentration of the naoh solution is ____.

Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
4 0

0.250 mol/L

<em>Step 1</em>. Write the chemical equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

<em>Step 2</em>. Calculate the moles of H2SO4

Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)

= 6.25 mmol H2SO4

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)

= 12.5 mmol NaOH

<em>Step 4</em>. Calculate the concentration of the NaOH

[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L

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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
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Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

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                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

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As reaction equation for the given reaction is as follows.

     NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

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Thus, we can conclude that volume of NaOH required is 20 ml.                    

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