Using data from Appendix D, calculate 3OH-4 and pH for each of the following solutions: (b) 0.035 M Na2S

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Hey everyone, we're asked to determine the concentration of hydroxide ions. The ph for a 0.105 moller potassium oxalate solution. And were provided R K A's first, let's go ahead and write out our reaction. We were told that we had potassium oxalate. Now this is going to disassociate into our potassium ions plus our oxalate ions. And in order to balance this out, we need a coefficient of two prior to our potassium. Now our potassium is a neutral ion and our Oxley ion is the conjugate base of our weak acid, which is oxalic acid. And as we can see from our reaction, we have a 1-1 ratio between our potassium oxalate and our oxalate ion. So this means that the concentration of our oxalate ion Is also going to be 0.105 moller. Now let's go ahead and create our ice chart. So we know that we had Oxley ion and this will react with water and our products are going to be our hydrogen oxalate. Since we're adding one proton plus our hydroxide ions. Now creating our ice chart. Initially we had 0.105 moller of our oxalate ion. We can completely disagreed our water since it is a liquid and we had zero of our products. Initially for our change we will have a minus X on our reactant side and a plus X on our product side. Since we're losing reactant and graining products at equilibrium in our react inside we have 0.105 minus X. And an X. And an X. In our product side. Since we are in our basic conditions, we have to find our KB. And as we've learned we know that our KB is equivalent to our K. W divided by R K. A. So plugging in those values, we know that our KW. 1.0 times 10 to the -14. And we're going to divide that by our K. A. Two which is 6.4 times 10 to the negative five. This gets us to a K. B. Of 1.5625 times 10 to the negative 10. Now to solve for X as we've learned our K B is equivalent to our products over our reactant. So we have an X times and X all divided by 0.105 minus x. Now, in order to check if the X. And our denominator is negligible, we can simply take 0.105 and divide it by R K B. Of 1.5625 times 10 to the negative 10. If we get a value that is greater than 500 then we can disregard our X. And in this case we do get a value greater than 500. So our X is negligible, solving for X. We get x squared is equal to 1.5625 times 10 to the negative Times 0.105 taking the square root of both sides. We end up with an X. of 4.05, 05 times 10 to the -6 Mohler. And this is going to be the concentration of our hydroxide ions. To further simplify this, we can write 4.1 times 10 to the -6 moller. Now to solve for a P. H. We can sulfur ph by looking at our P. O. H. We know that our P. O. H. Is equivalent to the negative log of the concentration of our hydroxide ions. So, plugging in these values, we get the negative log of 4.1 times 10 to the negative six. This gets us to a p. h. of 5.39. To solve for our p, all we need to do is take 14 and subtract 5.39 which gets us to a ph of 8.61. And this is going to be our final answers Now, I hope that made sense. And let us know if you have any questions

Using data from Appendix D, calculate 3OH-4 and pH for each of the following solutions: (b) 0.035 M Na2S

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Key Concepts

Dissociation of Sodium Sulfide (Na2S)

Hydroxide Ion Concentration (OH-)

pH Calculation