Determine the [OH^–] and pH of a solution that is 0.140 M in F^–.

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Hey everyone, we're told that a solution contains 0.650 moller of hipaa bro. My ion were asked to calculate the concentration of our hydroxide ions and the ph of our solution 1st. Let's go ahead and write out our reaction. We have our Hipaa bro my ion and we react this with water. When we react this with water, we're going to create hippo bromance acid and our hydroxide ions. Now let's go ahead and create our ice chart. We were told that we started off with 0.650 molar of our Hypo bro. Might. And we had zero of our products. Initially we can go ahead and disregard our water since it is a liquid and our change is going to be a minus X. In our react inside and a plus X. On our product side. Since we're losing reactant and gaining products at equilibrium on our react inside. We have 0.650 minus X. And an X. And an X. In our product side. Looking at our textbooks, we find that the K. A. Of hipaa brahmas acid Is going to come up to 2.8 times 10 to the -9. Since we are looking at hydroxide ions and our hypo bromide ion, we need the KB. Instead of the K. A. Now in order to find our KB we can go ahead and take our KW and divided by R. K. A. Plugging in our values, we have 1.0 times 10 to the negative 14 Divided by AK. A. of 2.8 times 10 to the -9. This will get us to a value of 3.5714 times 10 to the -6. Now let's go ahead and use our KB to solve for concentration of hydroxide ions. We know that our K. P. Is going to be equivalent to our products over our reactant. And again the reason why we used our K. P. Is because we are looking at a concentration of hydroxide ions. We can go ahead and check if we can disregard the X in our denominator By taking our 0.650 and dividing it by our KB Of 3. times 10 to the -6. And this will get us to a value that is greater than 500. So we can go ahead and disregard our X. If we had gotten a value less than 500 then we would have to keep our X. Simplifying our equation a bit further. We have 3.5714 times 10 to the negative six is equal to X squared over 0.650. Multiplying both sides by 0.650. And taking the square root of our both sides, We end up with an X. That 10 to 1.5- 3rd. And this is going to be the concentration of our hydroxide ions. And the concentration of our conjugate acid which is hipaa brahmas acid, solving for r p. O. H. We can then take the negative log Of 1.523, 6 times 10 to the negative third. And we end up with a p o. H of 2.82. Since we were asked for the ph We're going to take 14 and subtract 2.82. And this will get us to a ph of 11.2, which is going to be our final answer. So I hope this made sense and let us know if you have any questions.

Determine the [OH^–] and pH of a solution that is 0.140 M in F^–.

Verified Solution

Key Concepts

Acid-Base Chemistry

Equilibrium Constant (Kb)

pH and pOH Relationship

Determine the [OH–] and pH of a solution that is 0.140 M in F–.

Verified Solution

Key Concepts

Acid-Base Chemistry

Equilibrium Constant (Kb)

pH and pOH Relationship

Determine the [OH^–] and pH of a solution that is 0.140 M in F^–.