Question

500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. Will the resulting solutiomacidic,basic or neutral?Also calculate the molarity of the dilute solution.​

Answer 1

Answer:

Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M

Answer 2

"When 500 cc of 2N $Na_{2}CO_{3}$ is mixed with the 400 cc of 3N $H_{2} SO_{4}$ and volume was diluted to one liter then solution will be acidic and molarity will be 0.55 mol"

What is Molarity?

The molar concentration of a chemical species, specifically a solute in a solution, is measured in terms of the amount of material per unit volume of solution.

It is given that normality of $Na_{2}CO_{3}$ is 2 and $H_{2} SO_{4}$ is 3. It can be written in term of molarity like, the molarity of $Na_{2}CO_{3}$ is 0.1 M and molarity of $H_{2} SO_{4}$ is 1.5 M.

Now calculate the number of moles in $Na_{2}CO_{3}$

For $Na_{2}CO_{3}$ = 500 cc / 1000 cc/L ×0.1 mol/L =0.05 mol.

For $H_{2} SO_{4}$ = 400 cc/ 1000 cc/L × 1.5mol /L = 0.60 mol.

It can be said that 0.05 mol reacts with  0.05 mol $H_{2} SO_{4}$ .

Now calculate the molarity of  $H_{2} SO_{4}$ solution= 0.55 mol /L = 0.55 mol.

Hence, remaining 0.1 mol solution will make acidic.

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