Question

0.200M dilute hudrochloric acid was added to 25.0 cm3 of 0.400M calcium hydroxide solution until complete neutralization. What is the concentration of calcium chloride in the resulting solution?

Answer 1

@shubhamsrg

Answer 2

@ganeshie8 @mathslover would you kindly help me?

Answer 3

Calcium Hydroxide and hydrochloric acid reacts ib the following way:-

\[Ca(OH)_{2}+ 2HCl = CaCl _{2}+ 2H _{2}O\]

it is obvious that, no. of moles of calcium chloride produced will be equal to the number of moles of calcium hydroxide which reacts.
from this you can get the number of moles of calcium chloride and to calculate the concentration of calcium chloride, you need to divide the number of moles by volume of the new solution.
make sure that the method is clear to you..

1) to calculate the number of moles of calcium hydroxide,

recall that,
\[molarity = \frac{ number of moles }{ volume of solution ( L) }\]
molarity of calcium hydroxide = 0.4M (given)
and volume of solution = 0.025L
recall that 1L = 1dm^3 and 1mL= 1cm^3
this gives us.
no. of moles of calcium hydroxide = 0.01 moles..................

this is first part of the answer.
make sure that you understand this.
and if you find any difficulty, just ask. till then, i am writing the second part of the solution.

Answer 4

understood the method??

can you try the rest part now?

Answer 5

2) now, you need to find out the total volume of the solution, obtained by mixing 25mL of calcium hydroxide and an unknown amount of hydrochloric acid.

to get the new volume, you need to find out the unknown amount of hydrochloric acid added.

that you can calculate by calculating the amount of hydrochloric acid needed to completely neutralize 25mL of 0.4M calcium hydroxide.

no. of moles of calcium hydroxide = 0.01 mole. (from the first part of the answer.)
and we know,
\[Ca(OH)_{2} = Ca ^{2+} + 2OH ^{-}\]
this means 1 mole of calcium hydroxide will produce 2 moles of OH-
THUS, 0.01 mole of calcium hydroxide will produce 0.02 moles of OH- (simple unitary method)

and for complete neutralization to take place,

no. of moles of H+ = no. of moles of OH-
thus no. of moles of H+ = 0.02 moles.
and,
\[molarity = \frac{ no, of moles }{ vol. of solution (L) }\]

vol. of solution in litres = 0.02/0.2 = 0.1L or 100 mL

thus we added 100mL of HCl to 25mL of Ca(OH)2 for complete neutralization.
and thus, the total volume of new solution becomes 125mL. or 0.125L

the concentration of CaCl2 =\[\frac{ no. of moles of CaCl _{2} }{ volume of solution (L) }\]
thus, conc. = 0.01/0.125 = 0.08

Answer 6

is there some difficulty understanding??

Answer 7

thanks

Answer 8

welcome.!