How many ml of 0.250 M HCl would react exactly with 30.0 mL of the 0.150 M solution of Ca (OH ). solution ?
The chemical reaction involved is: HCl(aq) + NaOH (aq) NaCl (8q) + H20 (1) Given : Solution :

36.0 mL

Step-by-step explanation

The balanced chemical equation is: 2HCl_(aq) + Ca(OH)_2(aq) → CaCl_2(aq) + 2H₂O_(l)

1. Find the moles of Ca(OH)₂ by multiplying 0.0300 L with 0.150 mol/L Ca(OH)₂.

mol Ca(OH)₂ = 0.150 mol/L Ca(OH)₂ x 0.0300 L = 4.50 x 10^-3 mol

2. Convert mol Ca(OH)₂ to mol HCl using the 1:2 mole ratio of Ca(OH)₂ to HCl based on their coefficients in the balanced chemical equation.

4.50x10−3molCa(OH)2?(1molCa(OH)2?2molHCl?)=9.00x10−3molHCl

3. Convert mol HCl to L HCl using the given molarity of HCl (0.250 M).

9.00x10−3molHCl(0.250molHCl1L?)=0.0360L

4. Convert L HCl to mL HCl  using 1 L = 1000 mL as conversion factor.

0.0360L(1L1000mL?)=36.0mL

Therefore, 36.0 mL of 0.250 M HCl is needed to exactly react with 30.0 mL of the 0.150 M solution of Ca(OH)₂.