ACID-BASE NEUTRALIZATION REACTIONS. In this lab, we were given several vials whi

Question

ACID-BASE NEUTRALIZATION REACTIONS. In this lab, we were given several vials which each contained 5 mL of NaOH solution with a pH level of 9. We then added 1 mL of various HCl solutions with varying pH's to these vials.

Vial #1: 5 mL pH 9 NaOH solution
Vial #2: 5 mL pH 9 NaOH solution + 1 mL pH 13 HCl solution = 6 mL pH 12.13 solution
Vial #3: 5 mL pH 9 NaOH solution + 1 mL pH 11 HCl solution = 6 mL pH 3.12 solution
Vial #4: 5 mL pH 9 NaOH solution + 1 mL pH 5 HCl solution = 6 mL pH 6.55 solution
Vial #5: 5 mL pH 9 NaOH solution + 1 mL pH 3 HCl solution = 6 mL pH 6.95 solution
Vial #6: 5 mL pH 9 NaOH solution + 1 mL pH 1 HCl solution = 6 mL pH 1.81 solution

We're working with the chemical equation: HCl + NaOH <--> H2O + NaCl
Net Ionic chemical equation: H+ + OH- = H2O

Q) Calculate the number of moles of either OH- or H3O+ in the solution that was added (OH- if the solution was a base or H3O+ if the solution was an acid).

Explanation / Answer

For calculating number of moles ,first we have to calculate [H3O+] from Ph of solution

So for

Vial #1: 5 mLor 0.005L pH 9 NaOH solution

[H3O+]=10?pH=10?9=1.0?10?9

Now,calculate the number of moles from concentration

The number of moles of a substance in one litre of solution is called its molarity. The official symbol for molarity is “c” (concentration), but most people use the symbol “M”.

M = n/V,

where n is the number of moles and V is the volume in litres

So, n = 0.005 L soln × (1.0?10?9 mol H3O+/1 L soln) = 5*10-12 mol H3O+

Vial #2: 5 mL pH 9 NaOH solution + 1 mL pH 13 HCl solution = 6 mLor 0.006L pH 12.13 solution

[H3O+]=10?pH=10?12.13=7.4?10?13

n = 0.006 L soln × (7.4?10?13 mol H3O+/1 L soln) = 4.4*10-15 mol H3O+

Vial #3: 5 mL pH 9 NaOH solution + 1 mL pH 11 HCl solution = 6 mL pH 3.12 solution

[H3O+]=10?pH=10?3.12=7.5?10?04

n = 0.006 L soln × (7.5?10?04mol H3O+/1 L soln) = 4.5*10-6 mol H3O+

Vial #4: 5 mL pH 9 NaOH solution + 1 mL pH 5 HCl solution = 6 mL pH 6.55 solution

[H3O+]=10?pH=10?6.55=2.8?10?07

n = 0.006 L soln × (2.8?10?07mol H3O+/1 L soln) = 1.68*10-9 mol H3O+

Vial #5: 5 mL pH 9 NaOH solution + 1 mL pH 3 HCl solution = 6 mL pH 6.95 solution

[H3O+]=10?pH=10?6.95=1.1?10?07

n = 0.006 L soln × (1.1?10?07mol H3O+/1 L soln) = 6.6*10-10 mol H3O+

Vial #6: 5 mL pH 9 NaOH solution + 1 mL pH 1 HCl solution = 6 mL pH 1.81 solution

[H3O+]=10?pH=10?1.81=0.015488166

n = 0.006 L soln × (0.015488166mol H3O+/1 L soln) = 9.2*10-5 mol H3O+

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.