Calculate the pH and the concentrations of all species present
(H...

Question

Calculate the pH and the concentrations of all species present
(H2CO3, HCO3-, CO32-, H3O+ , and OH-) in 0.010 M
H2CO3 1Ka1 = 4.3 * 10-7; Ka2 = 5.6 * 10-112.

Accepted Answer

Hello. In this problem we have to find the concentrations of Susan IQ acid. It's mono printed form and its deep rooted form concentration of hydrogen ions and hydroxide ions in a 0.25 Mueller Susan IQ acid solution. And to determine the ph let's begin by making some shorthand notation. We call our acid H. Two A. The mono printed form H A minus and are deponent form a two minus. We'll make use of the two acid association equations and their equilibrium constant. To complete two ice tables will begin with the first one which is our acid undergoing hydraulic sis to form our mono product form of the acid. Cause I join your mines we have initial change and equilibrium initially we have 0. to 50 moller of our acid will ignore the water since it's a pure liquid and we have none of the monopoly native form of our acid or hydrogen mines are changes minus X plus X and plus X. The equilibrium is equal to our initial plus our change and writing our equilibrium cost expression Then have K. one is equal to concentration of our products. All over the concentration of our reactant water being a pure liquid does not appear in our equilibrium constant expression. Making use of the values from the ice table. We get x times x over 0. -1. And this we're told is equal to 6.17 Times -5. So we'll do a check to see if we can simplify this calculation. So to check are simplifying assumption. We're gonna take our acid concentration initially and divided by our equilibrium constant. The Sun works out to 0.0250, divided by 6.17 times 10 : -5. This works out to 405 which is not greater than 500. So we'll make use of the quadratic equation so we get X squared and is equal to 6. times 10 to minus five times 0.0 to 50 minus x. We'll simplify this further and bring everything over to the left hand side and so that is equal to zero. Then solve our quadratic equation. So X will be equal to negative B plus or minus the square root of b squared -4. AC. Oliver to a Make use of our quadratic equation. This then works out two X is equal to negative 6.17 times 10 to minus five plus minus the square root of b squared which is 6.17 times 10 to minus five squared minus four times a which is one times C which is negative, 1.54 to 5 times 10 to minus six. All over two times a. Which is one Solving for R. two possible routes. Yeah X is equal to 1.2115 Times 10 - Moller. And the other route is equal to negative 1.273, 2 times 10 to the -3 more. Since the death negative concentration doesn't make sense. We'll eliminate that one and so X. Then by looking at the ice table we see that this is equal to then the concentration of our mon opera naked form of our acid. And also the concentration of the my journey mines at equilibrium. And we can find the concentration of our acid From the ice table at Equilibrium sequel 0.0- -X. Which is 1.2115 times 10 - -3. So this works out to 0.0238 Molar. Moving on to our second equation. We then have our monopoly no protein ated form of our acid undergoing hydraulic sis form the deep protein, a form of our acid. Plus I journey mines we have initial change and equilibrium Based on the results from our last ice table. Our initial concentration of the monetary form of asset is 1.2115 times 10 -3. This is um also equal to our concentration initially of our hydro reminds we'll ignore the water since it's a pure liquid and then we initially have none of the deponent form of the acid. Our changes minus X plus X plus X. Will combine the initial and the change to get the equilibrium and then writing our equilibrium constant expression. We have the concentration of our products all over the concentration of our reactant water. Again being a pure liquid does not appear in our equilibrium constant expression, making use of the values from the ice table. We then get X times 1.2115 times 10 minus three plus X. All over 1.2115 times 10 to - minus X. And we're told that this is equal to 2.29 Times 10 : -6. So we're gonna check our simplifying assumption to see if we can remove the plus sex and the minus X. To do this. We're gonna take our initial monopoly ordinated acid concentration and divided by our equilibrium constant K. Two. This then works out to 1.2115 times 10-3, Divided by 2.29 times 10 : -6. And this works out to 529 which is greater than 500. So we can then simplify our calculation by eliminating the plus X and the -X. So we then get X Times 1.2115 Times 10 : -3. All over 1.2115 Times -3 is equal to 2.29 Times 10 : -6. So the value of 1.2115 times and my three will cancel in the numerator and the denominator And we're left with an X. is equal to 2.29 Times 10 to the -6 Mueller. From the ice table. This tells us that this is our concentration of our deep resonated acid. So we'll go back to our formulas and then given the X. Is so small, the concentration of the not a pro nated acid in the hydrating mind concentration be what we found from the first ice table. So 1.21 times 10 to minus three molar. And the concentration of or acid then an equilibrium will be equal to 2. 10 minus two. We also found. From solving the first ice table, we want to know the hydroxide iron concentration. So we can find this by taking our iron product constant of water and divided by the hydrogen ion concentration. My product Constant Water is one times 10 -14. And our high journey mind concentration with 1.21 understand minus three. So this works out to 8.25 times 10 to -12 Mueller. And the last thing we're asked to find was the ph so our ph is the negative log of our hydrogen mind concentration. Our ph works out to 2.92. So we have the concentration of our the person, a form of the acid mon opera in a form of the acid, which is also put to the concentration of our hydrogen mines and the concentration of our acid hydroxide iron concentration and our ph Of 2.92. Thanks for watching. Hope this helps.

Calculate the pH and the concentrations of all species present (H2CO3, HCO3-, CO32-, H3O+ , and OH-) in 0.010 M H2CO3 1Ka1 = 4.3 * 10-7; Ka2 = 5.6 * 10-112.

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