A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure
was...

Question

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure 
was dissolved in enough water to make 50.0 mL of 
hydrofluoric acid. 
(b) To what volume must you dilute the solution to triple 
the percent dissociation?

Accepted Answer

Hello in this problem, we're told a volume of 250 ml of gaseous hydrogen azide was recovered and isolated at 85°C and .470 atmosphere pressure. This recovered hydrogen gas was initially dissolved in water to make mL of hydrochloric acid solution In order to increase the percent organization % Organization 5 times to what volume does the solution need to be diluted forgiven the acid dissociation constant for hydrochloric acid. To begin, let's find the initial moles of hydrogen azide and then find the concentration initially of hydrochloric acid to be used in an ice table. So we are told that we have 250 ml of Hi genocide. We'll convert this from milliliters leaders And we are at a temperature of 85°C. So at 2 73. So our temperature will be 358 Kelvin. We can make use of the ideal gas equation to find the moles of hydrogen aside. So is equal to P. V. Over R. T. Our pressure is 0.470 atmospheres. Our volume is 0.25 leaders in our temperatures 358 Kelvin. We're going to make use of the r value of 0. leaders atmospheres for Calvillo. So our units of Leaders, atmospheres cancels and kelvin cancels who are left with moles. So our moles of I'm a guide or 0.0040 moles and now finding our concentration initially of hydrochloric acid. We put those 0.0040 moles of hydrogen azide into a 100 ml solution. Pervert our volume from milliliters, the leaders, our units and milliliters cancels. This works out then to 0. 00 moller. Now that we know our initial concentration that's generate a nice table. So use an ice table and our K. Value to find our hydrogen ion concentration at equilibrium. So we have one hydrochloric acid undergoing hydraulic sis form. Have you any minds and the conjugate base more acid. We have initial change and equilibrium our initial concentration as we found above 0.0400 for the hydrochloric acid will ignore water since it's a pure liquid. And initially we have none of the hydro any mines or our conjugate base are changes minus X. And then plus X. And plus sex will combine the initial and the change to get our equilibrium er equilibrium constant expression. Then we have a concentration of our products. All the concentration of our reactant water being a pure liquid does not appear in our equilibrium past expression. They can use the values in the ice table. We have then X times X over 0. minus X. And we're told that this is equal to 1.90 Times 10 to the -5. So we're going to check our simplifying assumptions to see if we can eliminate the -1. So we're gonna take our initial concentration of hydrochloric acid by the buyer. Pay value. And this works out to 2105 Which is greater than 500. So we can simplify our calculation and eliminate the minus X. In doing so. Then left with expert Over 0.0400 is equal to 1.90 clamps 10 -5. We will move the 0.040 to the other side and take the square root of both sides. This thing gives us X. is equal to 8.72 Time to tend the -4. Looking at the ice table. This then tells us concentration of hydrogen mines. So our percent organization can be found by taking our concentration of attorney mines divided by our initial concentration of our acid And multiplying by percent. Ization then is 2.179%. Our next step then we're going to determine volume of the solution needed to increase percent organization by five times so we take five times our percent organization. five times 2.179%. This gives us 10.897%. And so we'll convert this into a fraction related to previous information. So then this will be equal to then X. Which is the amount of irons that we form in solution divided by then our initial concentration of our acid. And so we know the moles. Um We know the moles of our acid. We need to determine the new volume. So we get 0.10897. Then it's equal to it's over 0.00400. Why? So I can solve for X. X 10 is equal to 0. times 0.400. Bye bye. Why? This works out to 4.3589 comes to the -4. Over Why I can now solve for Y. In terms of X. Y is equal to 4.35 8, 9 times 10 -4. Correct? Going to our definition of our equilibrium constant. You know, this is equal to 1.9 times 10 -5. It's equal to as we found with ice table X squared over our initial concentration of our acid minus X. Just focusing on the 0.00400 over Y. This animal be equal to 0. over 4.3589 times -4. Right? By X. Based on the relationship we developed between Y and X. This then is equal to 9.1766 x. You can then substitute this back into our equilibrium constant expression we had 1.9% of -5 is equal to x squared and over 9.1766 X. We've substituted for the 0.00400 divided by Y. This is then -X. Can solve for X squared x squared, then will be equal to 1.9 times 10 to - Times 8.1766 times x. This works out to 1.575, 4 Times 10 -4 times x. With by both sides by X. Let me then get X is equal to 1.575, 4 Times 10 : -4. And we can find why Why was equal to 4.30 to 5 times 10 to - provided by X, which is 1.5754 times 10 to minus four. This works out to 2. Leaders. So in conclusion then the solution should be diluted a volume 2.7 and Leaders to increase percent Organization 55 times. Thanks for watching. Hope this helped

A 1.000 L sample of HF gas at 20.0 °C and 0.601 atm pressure was dissolved in enough water to make 50.0 mL of hydrofluoric acid. (b) To what volume must you dilute the solution to triple the percent dissociation?

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