If 525 mL of 0.80 M HCl solution is neutralized with 315 mL of Sr(OH)2 solution what is the molarity of the Sr(OH)2? __ HCl + __ Sr(OH)2  __ SrCl2 + __ H2O

Question

Wyatt Atkins · Accepted Answer

The molarity of #Sr(OH)_2# is #"0.67 M"#

Since bases and acids cancel each other out, strychium hydroxide will neutralize hydrochloric acid to produce water and a salt.

#SrCl_2 + 2H_2O# -> #Sr(OH)_2 + 2HCl

Actions:

The quantity of #HCl# moles used during the procedure

# "Moles" = "Concentration" xxviii "Volume" #

is comparable to

#= "0.8 moles" / "1000 ml" * "525ml" = "420 millimoles" #

Here

#"milli" equals 1/1000#

The stoichiometry (or balanced equation) shows that half a mole of strontium hydroxide is neutralized for every mole of #HCl#.

Consequently, the quantity of strontium hydroxide in moles is equal to

Consequently, strontium hydroxide concentration will be

# "210 millimoles" / "315 ml" = "0.66667 millimoles/mL" = "0.67 moles/L" = "0.67 M" #

Xavier Avery · Answer

undefined $$ 2 \, \text{mol HCl} + 1 \, \text{mol Sr(OH)2} \rightarrow 1 \, \text{mol SrCl2} + 2 \, \text{mol H2O} $$

Molarity of Sr(OH)2: $ M = \frac{n}{V} $

Moles of Sr(OH)2: $ n = \text{Molarity} \times \text{Volume} $

$$ n_{\text{Sr(OH)2}} = (0.80 \, \text{mol/L}) \times (0.525 \, \text{L}) $$

Molarity of Sr(OH)2: $ M_{\text{Sr(OH)2}} = \frac{n_{\text{Sr(OH)2}}}{V_{\text{Sr(OH)2}}} $

$$ M_{\text{Sr(OH)2}} = \frac{n_{\text{Sr(OH)2}}}{0.315 \, \text{L}} $$

If 525 mL of 0.80 M HCl solution is neutralized with 315 mL of Sr(OH)2 solution what is the molarity of the Sr(OH)2? __ HCl + __ Sr(OH)2 __ SrCl2 + __ H2O

If 525 mL of 0.80 M HCl solution is neutralized with 315 mL of Sr(OH)2 solution what is the molarity of the Sr(OH)2? HCl + Sr(OH)2 SrCl2 + H2O