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Question

How many milliliters of .0215 M NaOH solution are needed to completely neutralize 2.50 mL of .0825 M H2SO4 solution?

Answer 1

idea is equivalents of NaOH (that is the number of OH's) = equivalents of H+'s from the H2SO4

here we go.....

2.50 ml H2SO4 x (1L / 1000 ml) x (.0825 moles H2SO4 / 1 L H2SO4) x (2 moles H+'s / 1 mole H2SO4) x (1 mole OH- / 1 mole H+) x (1 mole NaOH / 1 mole OH-) x (1 L / .0215 moles NaOH) x (1000 ml / 1 L)

= 2.50 x .0825 x 2 / .0215 ml NaOH = 19.2 ml

Answer 2

Since sulfuric acid produces 2 moles of H+ per mole of acid, you are neutralizing 2.5 mL of 0.1650 M monobasic acid. Since NaOH neutralizes this on a one-to-one basis:
V * 0.0215 = 2.5 * 0.1650 (V is NaOH mL)
V is about 20 mL

Answer 3

Moles H2SO4 = 2.50 x 0.0825 /1000 = 0.000206
Moles H+ = 2 x 0.000206 = 0.000412
Moles H+ = moles OH-
Molarity = moles / volume
volume = moles / Molarity = 0.000412 / 0.0215 =0.0192 L = 19.2 mL

Answer 4

This is a simple proportion, complicated only in that you need two moles of lye to neutralize one mole of acid.

Answer 5

form of like the Millikan oil drop test (seem it up). you ought to discover a small huge form that the mass of pennies could be a distinctive of. in case you had better than it is easy to with diverse numbers of pennies, this could be achievable, yet particularly difficult.