Question

In: Chemistry

A student is given 3 beakers: Beaker 1- 50.0 ml of a solution produced by dissolving...

A student is given 3 beakers:

Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak monoprotic acid ,HX, in enough water to produce 1 liter of solution. The empirical formula of HX is CH2O. The solution contains 3 drops of phenolphthalein.

Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

Beaker 3 – 50.0 ml of 0.250M KOH

The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.

Solutions

Expert Solution

Step 1, -

Since 50 ml of solution in beaker1 is completly reacted with 20.0 ml of 0.250 KOH, the molarity of solution in beaker 1 ; [ using relation M1 X V1 = M2 x V2 ]

.......................................................................................................................................= ( 20.0 x 0.250 ) / 50

...................................................................................................................................... = 0.10 M

Step 2, -

Calculate Molecular mass of HX ,assuming it as = 'M'

Since, 6.0 gms per litre solution of HX is = o.1M ..............calculated as above in step 1

Therefore a solution of HX containing one gm . molecular mass ( ie. 1M ) will contain = ( 6.0 / 0.1 ) = 60 .0 gms

Step 3, -

Given Emperical formula as CH2O

emperical formula wt = 30   

n = Mol. formula wt. / Emperical formula wt

.....=60/30 = 2

So , The acid is .......( CH2O )2 or ,...CH3COOH

X ie. anion in the acid is --- CH3COO

Calculation of pH-

The pH of the solution now would depend on the molarity of acidic solution due to hydrolysis of the salt formed as a result of following reaction,

......................................... CH3COOH + KOH = CH3COOK + H2O

Using the relation M1 x V1 = M2 X V2

the molarity of the NaX ( or in this case  is CH3COOK ) solution = ( 20.0 x 0.25 ) / 70

........................................................................................................ = 0.07M

which is also the molarity of NaX solution with pH 8.8 given in beaker 2

Thus the pH of beaker 1 after addition of 20 ml = 8.8

========================================================================================

-   


Related Solutions

A student is given 3 beakers: Beaker 1- 50.0 ml of a solution produced by dissolving...
A student is given 3 beakers: Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak  monoprotic acid ,HX, in enough water to produce 1 liter of solution.  The empirical formula of HX is CH O. The solution contains 3 drops  of phenolphthalein. Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8 Beaker 3 - 50.0 ml of 0.250M KOH The contents of beaker 3 is added drop-wise to...
You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contains...
You have 3 beakers: Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2 Beaker 2 contains 25.0 ml of 0.456 M NaC2H3O2 Beaker 3 contains 5.0 ml of 0.789 M HCl A.) What is the pH of the solution in beaker 1? B.) The contents of beaker 1 are poured into Beaker 2 and mixed. What is the pH of the resulting solution? C.) The contents of beaker 3 are now poured into the resulting solution. What is the new...
1. A student prepares 50.0 mL of a 2.5M solution by diluting 10.0 mL of a...
1. A student prepares 50.0 mL of a 2.5M solution by diluting 10.0 mL of a stock solution. a.) Calculate the concentration of the stock solution. b.) Calculate the volume of H2O used. 2. What is the volume of a 3.0M solution prepared by diluting 5.0 mL of 7.5M stock solution. How much water was used? Can you please show all of the work and calculations for #1 a&b and #2 so I can understand. Please and ThankYou!!!!
A solution was produced by dissolving NH3 to give a 250. mL solution with 3.00 M...
A solution was produced by dissolving NH3 to give a 250. mL solution with 3.00 M NH3. a. What is the pH of this solution? pH = b. What mass of NH4Cl must be added to this solution to give a pH of 9.45? __ g c. If 40. mL of 2.0 M KOH was added to this solution, what will be the pH of this solution? pH = d. If 40. mL of 2.0 M KOH was added to...
A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.300 M HCl....
A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.300 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.36 g of precipitate. Determine the concentration of the solution.
A student reaceted 50.0 ml of 1.05 N sodium hydroxide with 50.0 ml of 1.00 N...
A student reaceted 50.0 ml of 1.05 N sodium hydroxide with 50.0 ml of 1.00 N sulfuric acid in a coffee cup calorimeter. From the temperature vs time plot, the average Ti was 25.7 celsius and the Tf of the reaction mixture was 31.2 Celsius. (ignore calorimeter constant). What is the enthalpy of neutralization?
You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in...
You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 22.0C. The final temperature of the neutralization reaction was determined to be 22.5C. a) What is the total amount of heat evolved in this reaction? Show all work. b) If 0.135 moles of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (enthalpy) for this reaction?
50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both...
50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both at 20.0 degree Celsius, were mixed in a calorimeter. Calculate the molar heat of neutralization of HNO3 in kJ/mol if: (1) final temperature was 28.9 degree Celsius; (2) the mass of the overall solution was 102.0 g; (3) the heat capacity of the calorimeter was 25.0 J/C; (4) and assume that the specific heat of solution is the same as water, 4.184 J/(g C);
A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of...
A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is then added to the mixture. Assume no change in volume. Ksp = 2.2 × 10-20 for Cu(OH)2, Ksp = 1.3 × 10-15 for Co(OH)2 The hydroxide concentration at which the first metal hydroxide will just begin to precipitate is __________M Using the integer designators, the metal hydroxide that precipitates first is {(1) Cu(OH)2 or (2) Co(OH)2 } _____________M...
6. You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution...
6. You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 22.0OC. The final temperature of the neutralization reaction was determined to be 22.5OC. a) What is the total amount of heat evolved in this reaction? Show all work. b) If 0.135 moles of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (enthalpy) for this...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT