New data analysis page Data Analysis Calculate the average number of drops of HC
ID: 966919 • Letter: N
Question
New data analysis page Data Analysis Calculate the average number of drops of HCl used. Calculate the molarity of the Ca(OH)_2. This is an acid/base neutralization reaction. N_acidV_acid=N_baseV_base N=M_acid x (#H^+) N=M_base x (#OH^-) Volume can be recorded in drops. 2HCl + Ca(OH)_2 rightarrow 2 H_2O + Ca(Cl)_2 Determine the molarity of the calcium ions and hydroxide ion. Ca(OH)_2 rightarrow Ca^2+ + 2(OH)^- Calculate the K_sp = [Ca^2+] [OH^-]^2 Calculate the concentration of a saturated Ca(OH)_2 solution in grams/liter. Determine the concentration of a Ca(OH)_2 saturated solution in grams/100 mL.Explanation / Answer
The average of HCl would be:
13+11+11 / 3 = 11.67 mL
For the second question, let's calculate the concentration of Ca(OH)2:
The N of HCl is the same as molarity cause it has only 1 hydrogen so:
Nb = 0.1 * 11.67 / 2*40 = 0.015 N
Mb = 0.015 / 2 = 0.0075 M
Now, to get the concentration of Ca2+ and OH-
[Ca2+] = [CaOH2] = 0.0075 M
[OH-] = 2*0.0075 = 0.015 M
Ksp = 0.0075 * (0.015)2 = 1.68x10-6
We know that 0.0075 mol/L is the concentration, so in g/L we need the molecular weight which is 74 g/mol so:
C = 0.0075 mol/L * 74 g/mol = 0.555 g/L
C (g/100mL) = 0.555 g/L * 0.1 = 0.0555 g/100mL
Hope this helps