Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
162 views
in Chemistry by (90.6k points)
closed by
What volume of `0.00300 M Hcl` solution would just neutralize `30.0 mL` of `0.00100 M Ca(OH)_(2)` solution?
A. `20.0 mL`
B. `30.0 mL`
C. `78.60 mL`
D. `25.22 mL`

1 Answer

0 votes
by (88.6k points)
selected by
 
Best answer
Correct Answer - 1
Write the balanced equation for the reaction to determine the reaction ratio.
`2Hcl + Ca(OH)_(2) rarr CaCl_(2) + 2H_(2)O`
`2 mmol 1 mmol 1 mmol 2mmol`
Then (1) convert milliliters of `Ca(OH)_(2)` solution to millimoles of `Ca(OH)_(2)` using molarity as a unit factor, `0.00100 mmol Ca(OH)//1.00 mL Ca(OH)_(2)` solution, (2) convert millimoles of `Ca(OH)_(2)` to millimoles of HCl using factor, `2 mmol HCl//1 mmol Ca(OH)_(2)`. (the reaction ratio from the balance equation), and (3) convert millimoles of HCl to milliliters of HCl solution using the unit factor, `1.00 mL HCl//0.00300 mmol HCl`, that is molarity inverted.

`30.0 mL Ca(OH)_(2)xx(0.00100 mmol Ca(OH)_(2))/(1.00 mL Ca(OH)_(2))xx(2 mmol HCl)/(1 mmol Ca(OH)_(2))xx(1.00 mL HCl)/(0.00300 mmol HCl) =20.0 mL HCl`
Note that the balance chemical equation allows us to construct either a mole ratio or a millimole ratio
`(2 mol HCl)/(1 mol Ca(OH)_(2))` or `(2 mmol HCl)/(1 mmol Ca(OH)_(2))`
image

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...