Question 19125 mL of NH3 at0.105 MQuestion 20500.mLof KClat0.455 M...

If $29.50 \mathrm{~mL}$ of $0.175 \mathrm{M}$ nitric acid neutralizes $50.0 \mathrm{~mL}$ of ammonium hydroxide, what is the molarity of the base? $$ \mathrm{HNO}_{3}(a q)+\mathrm{NH}_{4} \mathrm{OH}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$

Hello. Today will be suffering for problems. 16 points a B. For part They were asked to calculate the morality of the ammonia solution. So if 29 point A. Of and each tree right mass Is Eco 200 g. That would mean the von solution equals 100 grands Divide that by zero hey 960 grands. Mm mary later grants cancer out. You get 11 point six. Yeah. Mm New leaves. Yeah, if you remember wowee Z equals two mass of NH tree. Mm And the more mass. So that would mean the twin nine he grams divided by the molar mass or Finance street. Mm So we're gonna put 70 g. Bye bye ball brands cancel L. And you modify that. Why amazon if I buy 11 16 Merrill leader and you're gonna get mm He was 15.7. Mhm. Now for part B. If you never previous video on the formulas when? Mm. Okay. Mhm. I don't know This first dam is 15 7 M. The first volume. Mhm. Is 300 10 male leaders. I don't know what the second concentration of this event is. And for the second volume is 2.5 Zoo leaders. So what you're going to do first is to convert this one into. We're all ears. Okay now you're going to plug and chug because there's no room to do right on. We're gonna write on here. So this becomes M. Two. The force 15.7. Yeah. Times your point leaders divided by five and you're gonna get one point 88 three negative three. That's our answer. In case you're wondering where I got the 0.3. I just converted this one. Sorry I just conferred 300 me released into leaders. Okay that's it for this video.

This is a pretty simple problem for this problem. We're going to find the concentration of ammonia. Here are givens. We have HCL, which were using to neutralize react with the ammonia. And we have, ah, volume of 39.47 milliliters. That's our volume. And we have a concentration of zero point 09 84 molars Moller for the ammonia. We're told that we have 50.0 mill leaders and we're trying to find the concentration. These two substances react in a 1 to 1 ratio so I can use the following equation the concentration of the acid times the volume of the asset equals the concentration of the base times the volume of the base, and we're gonna be solving for concentration debates. Concentration of the base is gonna equal the concentration. He asked that 0.984 polarity and I don't need to convert these two leader leaders because they're both of milliliters. So it'll work out perfectly. 39.47 mil leaders divided by 50.0 milliliters. And the concentration of my base is 0.7 seven seventh Moller. And that is ammonia problem done.

Mhm. Yeah. First of all, we need to determine the mass of solution, which is 25.8 g of ammonia multiplied with 100 g of solution divided by five g of ammonia. Yeah, Mass of the solution turns out to be 516 g of solution. That means There is 25.8 g of Ammonia present in 516 g of solutions. We know density is equal to mass per unit volume. So plugging in the values density is 1.00 six grams per milliliter With 516 g of solution mass and divided by volume. Volume turns out to be 512.92 millilitres. The amount of household ammonia needed to obtain 25.8 g of ammonia Will be 512 million later.

Here we are taking a look at the sala ability equilibrium for and I have 32. This is an equilibrium with and I two plus nickel in the plus the oxidation state at I. +03 minus where we've got to lots of those. So we can calculate the moles Is equal to the mass divided by the Molar Mass. We get not point not 1- two moles. So the amount of nickel present as the iron, It is not point not 1- two moles. So the I owe three to minus present is to lots of this. So we multiply it by two. We get not point out 244. Cause of the stroke geometric coefficient, The volume is one L. And so therefore our concentrations are not point not 1-2, no point not 244. And so to form a complex of N I N H 362 plus the amount of NH three required a six. Multiplied by not point Not 1- two moles. There's not point not 732 malls of NH three that we need. And so what we can do is write an expression for the K. C. Which is equal to and I and H three. We have two plus up here. Six down here, Multiplied by Bios three minor the power to because of the strike geometric coefficient divided by NH three are ligand to the power six which is equal to chaos. Multiplied by K sp. And so what we do is plug in values from our ice table including exes have not point not 1 to 2, multiplied by not point not to 44 squared divided by X. Take away not point not 732 squared. This is equal to 5.6 times 10 eight. Multiplied by 1.4 times 10 to the -8 is what we can do is solve for X. That is equal to not .1712. This is approximately not .17. We need not .17 miles of Ammonia need to be added to just dissolve the solid and I I Owe 3 2.