Determine the pH of each of the following solutions (Ka and Kb va...

Question

Determine the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (c) 0.165 M hydroxylamine.

Accepted Answer

Hey everyone we're us to calculate the ph of a 0.245 molar hydrogen solution and were provided the KB. Looking at our KB, we know that we have a weak base. So writing out our reaction, we have a hydrazine and this will react with water. Now our products are going to be the conjugate acid of hydrazine plus our hydroxide ions. Now to create our ice chart. Initially we had 0.245 molar of our hydrazine and we can go ahead and disregard our water since it is a liquid. And initially we had zero of our products formed for our change, it will be a minus X on our react inside since we're losing reactant and a plus X on our product side. Since we're gaining products at equilibrium on our react inside, we have 0.245 minus X. And an X. And an X. In our product side, solving for X. We can go ahead and use our kb of 1.3 times 10 to the negative six. And we know that our KB is equivalent to our products over our reactant. In this case it will be X times X. All over 0.245 minus X. Now we can go ahead and check if we can disregard our X. And our denominator by taking 0.245 and dividing it by R. K B of 1.3 times 10 to the negative six. If we get a value greater than 500 then we can disregard our X. And in this case we do so we can disregard our X. And our denominator solving for X. We have 1.3 times 10 to the negative six and this will be equal to x squared. We're going to go ahead and multiply 0.245 onto both sides. To eliminate 0.245 in our denominator taking the square root of both sides. We end up with an ex of 5.6436 times 10 to the negative four moller. And this is going to be the concentration of our hydroxide ions. Now to solve for R P. H, we can look for our ph by first solving for P. O. H. Which is equivalent to the negative log of the concentration of our hydroxide ions Plugging in those values. We get negative log of 5.6436 times 10 to the -4 moller. This gets us to a p. h. of 3.25. And since we are looking for our P. H, all we need to do is take 14 and subtract our P. O. H which is 3.25 which gets us to a ph of 10.75. And this is going to be our final answer. Now I hope that made sense. And let us know if you have any questions

Determine the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (c) 0.165 M hydroxylamine.

Watch next