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