Chemistry, asked by upendarnathyadav66, 1 year ago

A 25.0 mL sample of 0.10 M HCl is titrated with 0.10 M NaOH . What is the pH of the solution at the points where 24.9 and 25.1 mL of NaOH have been added?

(a) 3.70 , 10.70
(b) 3.30 , 10.30
(c) 3.70 , 10.30
(d) 3.0 , 11.0

Answers

Answered by BarrettArcher
32

Answer : The correct option is, (C)

Solution for part 1 :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of }HCl={0.10M}\times {0.025 L}=0.0025

\text{Moles of }NaOH={0.10M}\times {0.0249 L}=0.00249

NaOH+HCl\rightarrow NaCl+H_2O

As 1 mole of HCl combines with 1 mole of NaOH

0.0025 moles of HCl will combines with 0.00249 mole of NaOH

Thus NaOH is the limiting reagent as it limits the formation of products and HCl is an excess reagent.

And 0.00249 moles of NaOH will produce 0.00249 moles of HCl

Moles of HCl left= (0.0025-0.00249) = 0.00001 moles

The total volume of the solution will be

[tex]V _{total }= 25mL + 24.9mL = 49.9mL=0.0499L[/tex]

Now we have to calculate the concentration of H^+

\text{Concentration of }H^+=\frac{\text{Moles of }H^+}{\text{Total volume of solution}}=\frac{0.00001}{0.0499}=2\times 10^{-4}M

Now we have to calculate the pH.

pH=-\log [H^+]=-\log (2\times 10^{-4})=3.69

Solution for part 2 :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of }HCl={0.10M}\times {0.025 L}=0.0025

\text{Moles of }NaOH={0.10M}\times {0.0251 L}=0.00251

NaOH+HCl\rightarrow NaCl+H_2O

As 1 mole of HCl combines with 1 mole of NaOH

0.0025 moles of HCl will combines with 0.00251 mole of NaOH

Thus HCl is the limiting reagent as it limits the formation of products and NaOH is an excess reagent.

And 0.0025 moles of HCl will produce 0.0025 moles of NaOH

Moles of NaOH left= (0.00251-0.0025) = 0.00001 moles

The total volume of the solution will be

[tex]V _{total }= 25mL + 25.1mL = 50.1mL=0.0501L[/tex]

Now we have to calculate the concentration of OH^-

\text{Concentration of }OH^-=\frac{\text{Moles of }OH^-}{\text{Total volume of solution}}=\frac{0.00001}{0.0501}=1.9\times 10^{-4}M

Now we have to calculate the pOH.

pOH=-log [OH^-]=-\log (1.9\times 10^{-4})=3.72

Now we have to calculate the pH.

pH=14-pOH

pH=14-3.72=10.28

Therefore, the correct option is, (C)

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