Question

Part A

What volume of 0.105 M HClO4 solution is needed to neutralize 46.00 mL of 8.80×10−2 M NaOH?

Part B

What volume of 0.124 M HCl is needed to neutralize 2.88 g of Mg(OH)2? Part C If 27.0 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.755-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?

Part D

If 45.8 mL of 0.118 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Answer 1

Part A

Given:

[HClO4]=0.105 M , [NaOH] = 8.80E-2 M , volume of NaOH = 46.00 mL = 0.046 L

Solution:

We have to find the volume of HClO4. In order to get its volume we must find its moles by using mole ratio between NaOH and HClO4.

Reaction:

HClO₄(aq) + NaOH (aq) --- > NaClO₄ (aq) + H2O (l)

In this reaction we can see the mol ratio between HClO4 and NaOH is 1:1

We use the volume and molarity of NaOH to get its moles and by using the moles of NaOH we can get moles of HClO­₄.

n NaOH =volume in L x molarity = 0.046 L x 8.80E-2 M = 4.05E-3 mol NaOH

n HClO₄ = n NaOH x 1 mol HClO₄/ 1 mol NaOH =4.05 E-3 mol HClO4.

Calculation of volume of HClO₄

Volume of HClO­­₄ = Number of moles of HClO₄/ molarity = 4.05 E-3 mol / 0.105 M = 3.86 E-2 L = 38.6 mL

Ans : Volume of HClO₄ needed = 38.6 mL

Part B :

From the given mass of Mg(OH)₂ , we get its moles. And by using reaction stoichiometry we can get moles of HCl required to react with Mg(OH)­₂.

Mole and molarity of HCl is used to find its volume.

Reaction :

Mg(OH)₂ (aq) + 2HCl (aq) -- > MgCl2 (aq) + 2 H2O (l)

n HCl = n Mg(OH)₂ x 2 mol HCl / 1 mol Mg(OH)₂ ]

= (2.88 g / 58.3197 g per mol) x 2 mol HCl / 1 mol Mg(OH)­₂

= 0.098766 mol HCl

Volume of HCl = 0.098766 mol / 0.124 M = 0.7965 L = 796.5 mL

Part C

In part C we have to find molarity of AgNO3 and for that we calculate moles of AgNO3.

Reaction:

AgNO3 (aq) + KCl (aq) --- > AgCl (aq) + KNO3 (aq)

Moles of AgNO3 = mole of KCl x 1 mol AgNO3 / 1 mol KCl

= ( 7.55E-4 g / 74.5513 g per mol ) x 1 mol AgNO3 / 1 mol KCl

= 1.01 E-5 mol

Molarity of AgNO3 is calculated by using following formula.

Molarity = mol / volume

Molarity of AgNO3 = 1.01E-5 mol / 0.027 mL = 3.75 E-4 M

Part D

In part D we have to calculated mass of KOH and that is done by using moles of KOH.

Reaction :

KOH (aq) + HCl (aq) -- > KCl (aq) + H2O (l)

Mol ratio between HCl and KOH is 1 : 1

Mol KOH = ( .0458 L x 0.118 M ) x 1 mol KOH / 1 mol HCl

=0.005404 mol

Molar mass of KOH = 56.1056

Mass of KOH in g = 56.1056 g per mol ) x 0.005404 mol = 0.30 g