A9.96 ml sample of nitric acid required 13.25 ml of 0.085 m strontium hydroxide for titration. calculate the molarity of the acid solution. (hint: it's stoichiometry, you need the balanced equation)

The molarity of the acid solution = 0.226 M

Explanation:

Step 1: Data given

Volume of nitric acid = 9.96 mL = 0.00996 L

Volume strontium hydroxide = 13.25 mL = 0.01325 L

Molarity of strontium hydroxide = 0.085 M

Step 2: The balanced equation

2HNO3 + Sr(OH)2 → Sr(NO3)2 + 2H2O

For 2 moles HNO3 we need 1 mole of Sr(OH)2 to produce 1 mole Sr(NO3)2 and 2 moles H2O

Step 3: Calculate the molarity of the acid solution

n2 * C1 * V1 = n1*C2*V2

⇒ with n2 = the number of moles of Sr(OH)2 = 1 moles

⇒ with C1 = the molarity of HNO3 = TO BE DETERMINED

⇒ with V1 = the volume of HNO3 = 0.00996 L

⇒ with n1 = the number of moles of HNO3 = 2 moles

⇒ with C2 = the molarity of Sr(OH)2 = 0.085 M

⇒ with V2 = the volume of Sr(OH)2 = 0.01325 L

1*C1 * 0.00996 L = 2 * 0.085 M * 0.01325 L

C1 = (2*0.085*0.01325) / 0.00996

C1 =  0.226 M

The molarity of the acid solution = 0.226 M

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