Data was collected by students in an acid base titration lab. They used 1.63 M Ca(OH)2(AQ)
What is the molarity of the acid.

Data Table Two HNO3(AQ) Ca(OH)2(AQ)
Initial buret reading 37.5 mL 45.0 mL
Final buret reading 57.0 mL 67.8 mL

3.8 M

Explanation:

Volume of acid used VA= 57.0 - 37.5 = 19.5 ml

Volume of base used VB= 67.8 - 45.0 = 22.8 ml

Equation of the reaction

2HNO3(aq) + Ca(OH)2(aq) > Ca(NO3)2(aq) + 2H2O(l)

Number of moles of acid NA= 2

Number of moles of base NB= 1

Concentration of acid CA= ???

Concentration of base CB= 1.63 M

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA= CBVBNA/VANB

CA= 1.63 × 22.8 × 2/ 19.5 × 1

CA= 3.8 M

HENCE THE MOLARITY OF THE ACID IS 3.8 M.

your answer would be “dry”