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
Answer : The correct option is, (C)
Solution for part 1 :
First we have to calculate the moles of HCl and NaOH.
As 1 mole of combines with 1 mole of
0.0025 moles of will combines with 0.00249 mole of
Thus is the limiting reagent as it limits the formation of products and is an excess reagent.
And 0.00249 moles of will produce 0.00249 moles of
Moles of 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
Now we have to calculate the pH.
Solution for part 2 :
First we have to calculate the moles of HCl and NaOH.
As 1 mole of combines with 1 mole of
0.0025 moles of will combines with 0.00251 mole of
Thus is the limiting reagent as it limits the formation of products and is an excess reagent.
And 0.0025 moles of will produce 0.0025 moles of
Moles of 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
Now we have to calculate the pOH.
Now we have to calculate the pH.
Therefore, the correct option is, (C)