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Ch.17 - Aqueous Ionic Equilibrium

Chapter 17, Problem 33b

Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. b. 0.15 M NaF

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Hello. In this problem, we are asked to use an ice table to determine the ph of a 0.55 Mueller potassium hypochlorite solution were given the acid dissociation constant for hipAA florist acid, potassium hypochlorite with associate former potassium mines and hipaa quarry animals. Hipaa core. It is then the conjugate base of our weak acid hipaa chorus acid. So it will undergo hydraulic sis acting as a base. It will accept the proton from water, perform hipaa chorus acid and hydroxide ions in our reaction. Let's now set up an ice table. So initially we have 0.55 more of our hipaa coric annan will ignore the water since it won't appear in our overall equilibrium constant expression and initially we have none of the hipaa chorus acid or hydroxide are changes minus X. And then plus X and plus X. Our equilibrium is our initial change. So hypochlorite is a base. So we need a based association constant. You can find this by taking the iron product constant. Water divided by the acid dissociation constant. So 1.0 times 10 to minus 14. Fed by the acid dissociation constant for Hipaa forest acid. It's 2.9 times 10 to the -8. So this works out to 3.45 Times 10 to the - writing our equilibrium constant expression. Then we have our product concentrations provided by a reacting concentration. So the water again does not appear in our equilibrium constant expression. Using a nice table. This then works out the X times X Over 0.55 -X. We're then going to check to see if we can simplify our calculation to remove the minus X. So our check then we're gonna take the initial concentration. I have a hippo Coraline divided by K. B. This is 0.55 Divided by 3.45 times 10 -7. This works at 1.595 times 10 to the six, which is much larger than 500. So we are safe to then simplify our calculation. This then will work out two X squared over 0.55 after we drop the minus X. and we can solve for X. 10. Multiply both sides by 0.55 and take the square root. This then works out to 4.36 Times 10 to the -4. Based on the ice table extent is equal to our hydroxide ion concentration. So we can calculate the P. O. H. It's the negative log of our hydroxide iron concentration. That works out to then 3.36. We can then calculate the pP ph is equal to minus the P. O. H. That works out to 10.64. So the ph of our solution is 10.64. And this corresponds to answer d thanks for watching. Hope this help