Okay. So we've got a buffer solution here. Okay, it's made of ammonium ion and ammonia. So we've got the ph Ph is 10. Therefore the hydrogen ion concentration must be one times 10: -10 More. Okay. And we have the K. For the acid part of this which is the ammonium ion Which is 5.6 Time some -10. So we'll use our equation for buffers. H plus there's a K. A. Times the concentration of the acid which is ammonium ion divided by the concentration of our base which is ammonia. Since we have H plus in K. We'll go ahead and plug those in one Time. To turn the negative 10 Equals 5.6. I'm saying the negative 10 times are ratio. But so we're gonna go ahead and solve for that ratio now the ratio of our acid compared to our base and that's going to equal 0.18 21 mm. So we're gonna use that ratio now to answer some other questions. So in the second one we've given some, we've been given some information about ammonia. I know of 1.24 moller And the volume was .465 L. So I have 5, 77 moles Of the NH three. But we can use that ratio. We just found to answer some questions since concentration is moles per liter. And so we're doing moles per liter over moles per liter. Well, the leaders have to be the same. So we can also write this as the malls of ammonium ion divided by the moles of pneumonia, Equal that same ratio of .18. So we'll go ahead and plug in what we know here. Okay, we're looking for the moles of ammonia mayan and we'll divide that by the malls of ammonia. Set that equal to our ratio to our moles of ammonium ion. This problem Is going to be zero .10 moles. Okay, in our next example, Were given 2.08 g of ammonium ion. Well ammonium chloride. Right? So we'll go ahead and change grams to moles. This is ammonium chloride. So we use the molar mass, which is 53.5. So that will give us .0390 moles of ammonium chloride, which is the same as the ammonium. Bye on. And then we'll go ahead and use our ratio again. We've got moles of our acid divided by Moles of our base. So moles of NH three ph hasn't changed. So we need to keep this the same ratio So we can solve for moles of NH three here we'll get .216 Malls. But we've been asked for what volume of that. But they gave us the polarity. So if we divide by arm polarity here, .236 moles per liter. We divide that out. We'll get our leaders which is .915 L. Okay. Or 9.2 Times 10 of the Second Mill leaders about 920. So on our final example, we've got the polarity and the volume of our ammonium ion solution. So we'll go ahead and multiply more clarity times leaders to get moles. We've got .043 moles of ammonium ion NH four plus. So we'll go ahead. Use our same ratio moles of our acid Divided by moles of our base, which is NH three Most equal .18. So if we multiply that out, we'll get malls of ammonia that's going to come out to be .239 moles mm. And again they are asking us for the volume that we need. So we're gonna go ahead and divide it by the polarity Just .499 Moller, which will give us points 479 L, four, Times 10 of the Second Mill leaders if we want to use proper sig figs.