Question

1. Calculate the Molarity, NaOH(mol/L) for Standar 1, Standar 2 and Standard 3 with calculations

2. Write and balance the chemical reaction between NaOh and H2SO4

Molarity of standard H2SO4 solution (from the lab container) - 0.2411 M Write and balance the chemical reaction between NaOH and H2S04 (don't forget physical states): Standard 1 Standard 2 Standard 3 NaOH used (mL) (final – initial burette readings] 27.26m H2SO4 (mL) dispensed 15. Om 29.3lmu 16.0mL 31.14 ml 17.0ML Molarity, NaOH (mol/L)

Answer 1

The neutralization reaction between strong acid (H2SO4) an strong base (NaOH) is

H2SO4 + 2NaOH -----------> Na2SO4 + 2H2O

i.e 1 mol of H2SO4 will neutralize 1 mole of NaOH

Now for

i- Standard -1

Given concentration of H2SO4 solution taken = 0.2411 M

Volume of H2SO4 solution used = 15 mL

So moles of  H2SO4 used = concentration * volume

= 0.2411 M * 15 mL

= 0.2411 mol/ 1000 mL * 15 mL

= 0.0036 moles

That means moles of NaOH it neutralized = 0.0036 * 2

= 0.0072 moles

Again given volume of NaOH solution used = 27.20 mL

So Molarity of NaOH solution = moles of NaOH / Volume of NaOH solution in L

= 0.0072 moles / 27.20 mL

= 0.0072 moles / 0.02720 L

= 0.2647 moles/L

= 0.2647 M

Similalrly-

ii- Standard -2

Given concentration of H2SO4 solution taken = 0.2411 M

Volume of H2SO4 solution used = 16 mL

So moles of  H2SO4 used = concentration * volume

= 0.2411 M * 16 mL

= 0.2411 mol/ 1000 mL * 16 mL

= 0.00385 moles

That means moles of NaOH it neutralized = 0.00385 * 2

= 0.0077 moles

Again given volume of NaOH solution used = 29.31 mL

So Molarity of NaOH solution = moles of NaOH / Volume of NaOH solution in L

= 0.0077 moles / 29.31 mL

= 0.0077 moles / 0.02931 L

= 0.2627 moles/L

= 0.2627 M

And

iii- Standard -3

Given concentration of H2SO4 solution taken = 0.2411 M

Volume of H2SO4 solution used = 17 mL

So moles of  H2SO4 used = concentration * volume

= 0.2411 M * 17 mL

= 0.2411 mol/ 1000 mL * 17 mL

= 0.00409 moles

That means moles of NaOH it neutralized = 0.00409 * 2

= 0.00818 moles

Again given volume of NaOH solution used = 31.14 mL

So Molarity of NaOH solution = moles of NaOH / Volume of NaOH solution in L

= 0.00818 moles / 31.14 mL

= 0.00818 moles / 0.03114 L

= 0.2626 moles/L

= 0.2626 M