What is the pH at 25 C of water saturated with CO2 at a partial pressure of 1.10 atm? The Henry's law constant for CO2 at 25 C is 3.1 * 10-2 mol>L@atm.

Video transcript

Hello everyone today, we have the following problem. Soda water is saturated with C. 02 at a partial pressure of 5.25 atmospheres at 25 degrees Celsius, calculate the ph of the soda water at 25 degrees Celsius. Given that Henry's Law constant for C. 02. At this temperature is 3.1 times 10 to the negative second moles over liters times atmospheres. And then we are given two K. A. Values for bicarbonate. So Henry's law for the situation is the soluble itty of C. 02 is equal to or directly proportional to Henry's constant times the partial pressure of CO. Two. And so we have our values. We have our K. Value here Which is our 3.1 Times 10 to the negative second moles over leaders times atmospheres. And then we have our partial pressure for C. 02 which is a 5.25 atmospheres. And when our units cancel out, we are left with 0. moles over L or Polarity. Second we want to write out our chemical reaction. So we have our C. 02 in the gaseous form combining with liquid water to form our carbonate, R C H two, C 03 in the acquis form. And so therefore we can say that the concentration of our C. 02 is equal to our concentration of our carbonate, Which is therefore going to be equal to our 0. charity. We can then construct our ice table. So we're going to ride out this equation here, our carbonate, H 2, 3. Aquarius plus our water will yield us our bicarbonate ion plus hi joanie, um So we have our initial are changing concentration and our equilibrium. So our initial concentration of our carbonate is 0.16275. Polarity water is not counted in this table because it is a constant, It is going to be zero for both the bicarbonate and the hydro liam. Our change is going to be -1. Since that's what we are taking away and we are adding it to this side here so that both of these reactors will have a positive x. So the equilibrium will be 0. minus X. And the equilibrium for both of these reactant or product on the right will just be X. And so we have our two K A. Values that were given to us. And we can note Our special make special notice that RKK one is greater than our KA two. And so this is what we will be using for this question. So if you set our expression up, this will be our fifth step here We have our KA one is equal to our equilibrium expression which will be our bicarbonate times. Are hydro ni um over our concentration of our carbonate. And so how do we get this While we took our concentrations of our products here, we multiplied those together and then we place them over our reactant. And so since water is a liquid, we do not count liquids in our expression. And so we can now solve a little bit more and go in depth with this equation here. So our KA one is 4.3 times 10 to the -7. And for our bicarbonate and our hydro ni um we can recognize those as X. So this will end up being X squared Over 0. minus X. Or a concentration of carbonate. The next thing that we want to make notice of is what is our K. A. Value. So our standard here would be 500. And so if we were to take our concentration of carbonate, our initial concentration and place it over our K. A. Value that we're using, Which would then be equal to 0. over our 4.3 times 10 to the negative seventh. This would be greater than 500. And so therefore x can be ignored since it is by far very large. And so we then just end up with 4.3 times 10 to the negative seventh equals X squared over 100.16275 without the X. So this is the equation that we will use solving for X in step seven, we get that X is equal to our concentration of hydro ni um Which is 2.6454 times 10 to the negative 4th polarity. So we're going to box that in. And so essentially we are looking for the ph so with our concentration of hydro ni um we can easily find our ph and ph is going to be equal to negative log times our hydro liam concentration, which will be negative log times 2.6454 times 10 to the negative four polarity. And this will give us a ph of 3.58 as our final answer. Overall, I hope this helped and until next time.