If 16.3 mL of a 0.185 M Sr(OH)2 solution is used to titrate the HCl in a 25.0-mL (0.0250 L) sample of gastric juice, what is the molarity of the HCl solution ?2HCl(aq) + Sr(OH)2(aq) → 2H2O(l) + SrCl2(aq)

Question

If 16.3 mL of a 0.185 M [latex]Sr(OH)_{2}[/latex] solution is used to titrate the HCl in a 25.0-mL (0.0250 L) sample of gastric juice, what is the molarity of the HCl solution ?

[latex]2HCl(aq) + Sr(OH)_{2}(aq) → 2H_{2}O(l) + SrCl_{2}(aq)[/latex]

Accepted Answer

STEP 1 State the given and needed quantities and concentrations.

ANALYZE THE PROBLEM	Given	Need	Connect
	2S.O ml (0.0250 L) of HCI solution, 16.3 ml of 0.18S M [latex]Sr(OH)_{2}[/latex] solution	molarity of the HCl solution	molarity, mole-mole factor
	Neutralization Equation
	[latex]2HCl(aq) + Sr(OH)_{2}(aq) → 2H_{2}O(l) + SrCl_{2}(aq)[/latex]

STEP 2 Write a plan to calculate the molarity.

[latex]\begin{array}{l}mL of Sr(OH)_{2} \ ext{solution}\end{array} \boxed{\begin{array}{l} ext{Metric}\ ext{factor}\end{array}} → \begin{array}{l}L of Sr(OH)_{2} \ ext{solution}\end{array} \boxed{Molarity}→ \begin{array}{l} moles of \ Sr(OH)_{2}\end{array} \boxed{\begin{array}{l} ext{Mole-mole}\ ext{factor}\end{array}}→ \begin{array}{l} moles of\ ext{HCl}\end{array} \boxed{\begin{array}{l} ext{Divide by}\ ext{liters}\end{array}}→ \begin{array}{l} ext{molarity of }\ ext{HCl solution}\end{array} [/latex]

STEP 3 State equalities and conversion factors, including concentrations.

[latex]\begin{array}{r c}\boxed{\begin{matrix} 1000 mL of Sr(OH)_{2} solution = 1 L of Sr(OH)_{2} solution \\frac{1000 mL Sr(OH)_{2} solution}{1 L Sr(OH)_{2} solution} ext{ and } \frac{1 L Sr(OH)_{2} solution}{1000 mL Sr(OH)_{2} solution} \end{matrix}} & \boxed{\begin{matrix} 0. 185 mole of Sr(OH)_{2} = 1 L of Sr(OH)_{2} solution\ \frac{0. 185 mole Sr(OH)_{2}}{ 1 L Sr(OH)_{2} solution} ext{ and }\frac{ 1 L Sr(OH)_{2} solution}{0. 185 mole Sr(OH)_{2}} \end{matrix}}\end{array}[/latex]

[latex] \begin{array}{r c}\boxed{\begin{matrix} 2 moles of HCl = 1 mole of Sr(OH)_{2} \\frac{2 moles HCl}{1 mole Sr(OH)_{2}} ext{ and } \frac{1 mole Sr(OH)_{2}}{2 moles HCl} \end{matrix}} \end{array} [/latex]

STEP 4 Set up the problem to calculate the needed quantity.

[latex] 16.3 \cancel{mL Sr(OH)_{2} solution} imes \boxed{\frac{1 \cancel{L Sr(OH)_{2} solution}}{1000 \cancel{mL Sr(OH)_{2} solution}}} imes \boxed{\frac{0.185 \cancel{mole Sr(OH)_{2}}}{1 \cancel{L Sr(OH)_{2} solution}}} imes \boxed{\frac{2 moles HCl}{1 \cancel{mole Sr(OH)_{2}}}} = 0.00603 mole of HCl [/latex]

[latex] molarity of HCl solution = \frac{0.00603 mole HCl }{\boxed{0.0250 L HCl solution }} = 0.241 M HCl solution [/latex]

Question 11.12: If 16.3 mL of a 0.185 M Sr(OH)2 solution is used to titrate ...

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