Question

Calculate the molarities and normalities of the solutions obtained on mixing 100 mL of 0.1 M $\sf H_2 SO_4$ with 50 mL of 0.1 M NaOH.
[Ans: Normality = 0.1 N, Molarity = 0.05 M]​

Answer 1

Answer:

right answer normality 0.1

Answer 2

GIVEN :

$\bullet$ Molarity of $\sf H_2SO_4$ = 0.1 M

$\bullet$ Volume of $\sf H_2SO_4$ = 100 ml

$\bullet$ Molarity of NaOH = 0.1 M

$\bullet$ Volume of NaOH = 50 ml

$\sf \bold{To\:find\::}$

$\bullet$ Molarity of resulting solution

$\bullet$ Normality of resulting solution

$\sf \bold{Formula\:used\::}$

$\bullet$ Molarity = Number of moles of Solute ÷ Volume of solution ( in L)

$\bullet$ Normality = Number of Equivalent mole of solute ÷ Volume of solution ( in L)

$\bullet$ Normality = Molarity × n-factor

$\sf \bold{Solution\::}$

a) First of all , here we are given one acid and one base . The acid and base will react to form salt.

$\sf H_2SO_4 \:+\: 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

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b) Now we have to find , moles of $\sf H_2SO_4$ , and NaOH

➝ Moles of $\sf H_2SO_4$ = Molarity of $\sf H_2SO_4$ × Volume of $\sf H_2SO_4$ (in L)

➝ Moles of $\sf H_2SO_4$ = 0.1 × 100/1000

➝ Moles of $\sf H_2SO_4$ = 10/1000 = 10 milimole

➝ Moles of NaOH = Molarity of NaOH × Volume of NaOH (in L)

➝ Moles of NaOH = 0.1 × 50/1000

➝ Moles of NaOH = 5 milimole

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c) Now , we will find limiting agent

➝ According to balanced reaction, 2 mole of NaOH will react with 1 mole of $\sf H_2SO_4$

➝ Therefore 1 mole of NaOH will require = (1/2) mole of $\sf H_2SO_4$

➝ Therefore 5 milimole of NaOH will require = (1/2)×5 milimole of $\sf H_2SO_4$

This gives that NaOH is limiting reagents

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d) Find moles of $\sf H_2SO_4$ & NaOH left

➝ As we know NaOH is limiting reactants , therefore moles of NaOH left = 0

➝ Moles of $\sf H_2SO_4$ left = Moles of $\sf H_2SO_4$ initially present - Moles of $\sf H_2SO_4$ that took part in reaction

{ From above we know that ,

$\bullet$ Moles of $\sf H_2SO_4$ initially present = 10 milimole

$\bullet$ Moles of $\sf H_2SO_4$ that took part in reaction = 5/2 = 2.5 milimoles }

➝ Moles of $\sf H_2SO_4$ left = 10 - 2.5 milimole

➝ Moles of $\sf H_2SO_4$ left = $\sf \bold{7.5\: milimole}$

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e) Find out final volume of solution

➝ Final volume = Volume of NaOH + Volume of $\sf H_2SO_4$

➝ Final volume = 50ml + 100ml

➝ Final volume = $\sf \bold{150\:ml}$

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f) Molarity of resulting solution = Moles of $\sf H_2SO_4$ left ÷ Volume of resulting solution (in L)

$\sf \implies Molarity \:of \:resulting\: solution \:= \: \dfrac{7.5 \times 10^{-3}}{150 \times 10^{-3}}\\\\\sf \implies Molarity \:of \:resulting\: solution \:= \: \dfrac{7.5}{150} \\\\\sf \implies Molarity \:of \:resulting\: solution \:= \: \bold{0.05\:M}$

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g) Normality of resulting solution

$\bullet$ First of all we need to find n-factor

➝ As we know resulting solution contain $\sf H_2SO_4$ not NaOH.

➝ n-factor of resulting solution = n-factor of $\sf H_2SO_4$

➝ n-factor of resulting solution = 2

$\sf \implies Normality\: resulting \: solution \: = 2 \times 0.05\\\\\sf \implies Normality\: resulting \: solution \: = \bold{0.1\:N}$

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$\sf \bold{ANSWER\::}$

$\sf \bullet \:\:Molarity\: = \bold{0.05\:M}\\\\\sf \bullet \:\: Normality \: = \bold{0.1\:N}$