Chemistry question about solutions.

Question

I don't manage to solve this problem, can someone help me?

 "Calculate how many cm^3 of a solution of sulfuric acid 80.0% (density = 1.727 g/mL) must be diluted with water in order to prepare a solution of 0.2 L 0.4 M. What's the pH of this solution if we add 800 cm^3 of water and 6g of NaOH?".

Thank you very much!

Answer 1

First you need to calculate the concentration of 80% solution of H₂SO₄. It contains 80 g of the acid per 100 g of solution. Taking into account the density of 1.727 g/mL, 100 g of the solution corresponds to 100/1.727=58 mL (0.058 L) of the solution. Molecular weight of H₂SO₄ is 2*M(H)+M(S)+4*M(O)=2*1+32+4*16=98 g/mol. Therefore, 80 g corresponds to 80/98=0.816 mol. Now you can calculate the concentration of H2SO4: 0.816 mol/0.058 L = 14.1 M. To prepare 0.2 L of 0.4 M solution, 0.2L*0.4 mol/L=0.08 mol of H₂SO₄ is needed. this amount of moles is in 0.08/14.1=0.0057 L (or 5.7 mL). therefore, 5.7 mL of 80% acid solution need to be diluted with 200-5.7=194.3 mL of water.

Second, when 6g of NaOH is added with 800 mL of water, some acid becomes neutralized. The equation is

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O. From the equation it is known that 2 moles of NaOH are needed for neutralization of 1 mole of the acid. Correspondingly, 0.16 moles of NaOH are needed to neutralize 0.08  moles of H2SO4. 6g of NaOH corresponds to 6/40=0.15 moles (40 g/mol is molecular weight of NaOH). It is enough to neutralize only 0.075 moles of the acid, and 0.08-0.075=0.005 moles of H2SO4 is left in solution. Each mole of H₂SO₄ dissociates into 2 moles of H⁺. 0.005 moles of the acid results in 0.01 moles of H⁺. After addition of 800 mL of water, the solution volume becomes 200+800=1000 mL (or 1L). Therefore, the concentration of H⁺ in the solution is 0.01M. The pH of the solution is -lg([H⁺])=2.

Answer 2

  Hi Angelica from New York, N.Y.

First question:

This type of calculations require two steps. first the concentrations of all solutions  , to be comparable, must be reduced to the same units. Second. calculate the quantity available and use the necessary amount to satisfy the question.

 

Since the question requires a molar concentration ,we will convert the % concentration in molar concentration.80% means 80g of acid present in 100g of solution. we can calculate the number of moles 80g represent by dividing by the gram mole of H2SO4 that is 98 ( S+ O4+H2= 32+68+ 2= 98).

100g of solution contain 80/98 = 0.816 moles of acid. Since the molarity is expressed per liter of solution we must convert 100g of solution to its volume. We know the specific gravity is 1.727  hence from the definition of density the volume of 100g of solution is given by: V=100/1.727=57.9 ml. Now we know that 0.816 moles are dissolved in 57.9 ml of solution ,therefore the number of moles in 1000 ml(a liter) is given by ( 0.816/57.9) x 1000 = 14.09 M

Now we have a 14.09M  H2SO4 solution and we want to dilute it with distilled water to prepare 0.2L  (200ml) of 0.40M acid solution.  200ml of this solution contain ( 0.4/1000 ) x200  = 0.08 moles of acid.

We must get this amount of acid from the concentrated solution and dilute it to 200ml.

 the conc.acid contains 14.09 moles per liter hence 14.09/1000= 0.08/X  (  X=ml of conc. acid to be used for the dilution ).

               14.09 : 1000 = 0.08 : X

 

and              14.09 X = 0.08 x 1000

and   X = 5.68

 

 And the final answer : 5.7 ml of the concentrated acid are diluted to 200ml with distilled water to obtain a 0.4M solution. The same result is obtained using the formula for calculating mixture properties used to answer the question of Addie from Conneaut Lake PA concerning the preparation of a metal alloy.

 

Now the pH question.

To 200ml 0f 0.4M solution H2SO4 are added 6g of NaOH. in accordance with:

      H2SO4 + 2NaOH = Na2SO4 + 2H2O

sulfuric acid is a divalent acid ( see my answer to Brieonnas's question on neutralizing H3PO4 with

NaOH ) therefore its gram equivalent (Eq ) is given by M/2  and our acid solution is 0.8 normal ( 0.8 Eq/L ) and NaOH is a monovalent base with Eq =M

 

Therefore 6g of NaOH are  6/40 =0.15 Eq of base.   This , by definition is also the number of Eq of acid that react completely with it .  This means that 0.15Eq of acid are neutralized. Since in 200ml of acid there are (0.8/1000) x 200 =0.16Eq of acid and 0.15 Eq are neutralzed by the base ,the remaining acid is only 0.16-0.15 =0.01 Eq in 200ml . When we add 800ml of water,the same number of equivalents will be in a liter of solution ( If we disregard volume changes since the concentrations are low). the final [ H+ ] concentration will be 0.01

and the pH =-log [H+ ] =  log 10^-2 = 2.

 

kind regards.

Ed Canelli