Enter Tour ansrerhe provided box. dilute nitric acid solution Fo' rrncts Wth thioc}- nnn (SC) lorm dark red compler: [Fe(H,O)a] ' - SCN =H,o (Fe(H,O)NCS]&...

In a dilute nitric acid solution, $\mathrm{Fe}^{3+}$ reacts with thiocyanate ion ( $\mathrm{SCN}^{-}$ ) to form a dark-red complex: $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{SCN}^{-} \rightleftharpoons$ $\mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$ The equilibrium concentration of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$ may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, $1.0 \mathrm{~mL}$ of $0.20 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$ was mixed with $1.0 \mathrm{~mL}$ of $1.0 \times 10^{-3} \mathrm{M} \mathrm{KSCN}$ and $8.0 \mathrm{~mL}$ of dilute $\mathrm{HNO}_{3} .$ The color of the solution quantitatively indicated that the $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$ concentration was $7.3 \times 10^{-5} M$. Calculate the formation constant for $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$

For a Let's calculate the P. Um, we have point 20 grams of sodium oxide, and let's convert this two moles. A molar mass is 61.98 grams being one more. This would yield 3.2 times 10 to the negative three moles and the polarity of sodium oxide is 3.2 tubs, 10 to the negative. Three moles over 0.1 User leaders. This would yield uh, 0.32 Moeller. Uh, let's disassociate this here in a 20 plus each Teoh would yield two in a plus, two of which minus went 032 Moeller 0.32 Moeller So free 30 plus, which would be acquitted. K w over which k w over the O. H minus each 30 plus works of to, uh, actually sorry. This is a 1 to 2 more ratio. This would be 0.64 uh, 0.64 0.64 this workout to 1.6 times 10 to the negative 13. And for the ph would be cool to, uh, negative log of each three go plus, which is negative. Log 1.6 times 10 to the negative 13 in itself for the pH. Here. When we get a ph of 12 point eat zero for a for part B, we have, uh, 1.26 grams of general three. It's convert this to Moules Moeller Mouse Nature gas ID is 63 points. Your grams per mole in this one year old went 0 to 00 moles, and then the polarity of the nectar kassid is 0.200 moves over the point five years your leaders and we will get 0.400 Moeller Nitric acid is a strong acid, 30 in 03 minus 0.4 0.4 and therefore the pH. Here's negative. Log age three plus negative log 0400 and we find a pH. That is 13 um, nursery one point 40 There's the pH of solution. Be for a solution. See, we're going to have a solution here. So, um, of a point 075 Moeller of the barium hydroxide is deleted. So uh, 0.4 leaders and point three easier zehr leaders that needed value becomes 30.10 Molder Barium hydroxide undergoes association. Be a two plus and 20 H. Minus 0.10 muehler 1 to 2 more issue and it's all free. Three or plus Kate of you over over H minus 0.0 times 10 to the negative. 14 over a 0.20 And we get 5.0 times 10 to the negative. 13. Moeller. That's fine. PH here is negative. Log 30 just negative log 5.0 times 10 to the negative 13 and we confined that the pH here is going to be pulled the 12.30 Lastly for D. We've got, uh, mixture of HCL in each and three, and so start with each CEO strong acid. It's too. His three plus c o minus and its CEO is at 0.20 Moeller, therefore, Eastern Plus would be put to use through Moeller 1 to 1, um, more racial and each of the three h 20 page 30 plus hand 03 minus went by. Zero Moeller 50 Moeller. So it's 30 NH 30 So the beach 30 plus in food off would be equal to No, we're going. We're told that the volumes are additive. So this would g, um went to zoom ALS plus 0.50 moles over one plus one leaders. This would work out to immaturity of 10.35 Moeller on through a P H is equal to negative log H 30 plus, which is a negative log 0.35 we confined that are pH Would because the point 45

So in this video, we're gonna go over Question 44 from Chapter four, which says a solution was prepared by mixing 50 milliliters of 500.1, Mueller 18 03 and 100 leaders of Point to Mueller H and 03 calculate the polarity of the final solution of nitric acid. So basically, polarity is concentration and units of most per leader. So on. In order to calculate the polarity of the final solution of nitric acid that results from mixing or two solutions, we need to know the number of moles and solution one and the number of Molson solution to and the blame of solution one in the volume of solution to and then our concentration will be the sum of the number of moles divided by the sum of the volumes. Eso First, let's figure out how many moles of nitric acid we have in the our first solution. So we take our point. 05 liters are 50 milliliters and we multiply a buyer 500.1 miles per leader and that gives us 0.5 moles. And then in our second solution, we take our 100 mill liters or 1000.1 leaders times are 0.2 moles per leader, and that gives us 0.2 bowls. And then our concentration of agent 03 is just in one plus in two, divided by V one plus V, too. So we have 10.5 point zero to that gives this 00.25 moles. And then we have 0.5 leaders plus 0.1 leaders. That gives this 0.15 leaders on dhe. This divides out 2.167 Moeller. So that's our concentration of nitrogen.

Okay, so we have a starting 1.15 leaders of our each and every it's most implied at by similarity that 0.100 moles over one meter multiply that by it. Um, more math. 63.12 eight grams over one moment and then multiply that by 100 grams of our solution over 70.3 grams of our age and all three and last. We must find out by one no leaders of our solution over 1.41 grams of our solution That gives us 7.31 milliliters of our solution.

In this problem and I cost solution containing as to minus iron will give it will give it will give I'm just writing the answer so just look at it carefully. It will give yellow Police see Pete dead with the suspense in all c d c 03 in water solution. Also, it will give black precipitated black precipitated with lead as he did solution. So according to the option obstinacy and d. R. Correct here.