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Lette (rciy L ' K| (3f 44 6 (S*202 C + 24_ < Ch4 ( revise LT > 0n0 nihpiyk 21 (1 0_+2c+N 6>2HCN(xr K3 Ka K(k)" optlon D AB solution initiall Cons...

Question

Lette (rciy L ' K| (3f 44 6 (S*202 C + 24_ < Ch4 ( revise LT > 0n0 nihpiyk 21 (1 0_+2c+N 6>2HCN(xr K3 Ka K(k)" optlon D AB solution initiall Consider the acid dissociation of chlorous acid as described by thereaction beloyy: If the commond established? initially contains 0.0S00 M HCIOz(aq), what is the concentration of HCIOz once equilibrium is HCIOz(aq) - Kc = L.0x 102 cE HO() = CIOz (aq) + H,Ot(aq) tabk 0.0204 M 0. 0500 0,0314 M 0.0186 M 0.0265 M 10.0390 M 0 Kio 4.5 ( Cosc +

Lette (rciy L ' K| (3f 44 6 (S*202 C + 24_ < Ch4 ( revise LT > 0n0 nihpiyk 21 (1 0_+2c+N 6>2HCN(xr K3 Ka K(k)" optlon D AB solution initiall Consider the acid dissociation of chlorous acid as described by thereaction beloyy: If the commond established? initially contains 0.0S00 M HCIOz(aq), what is the concentration of HCIOz once equilibrium is HCIOz(aq) - Kc = L.0x 102 cE HO() = CIOz (aq) + H,Ot(aq) tabk 0.0204 M 0. 0500 0,0314 M 0.0186 M 0.0265 M 10.0390 M 0 Kio 4.5 ( Cosc +0 on Solution: CcIo-JCHo ] 4C1oz Icioz Hz0 C4cioz] 0500 CxJCx] Co,0soo - X 0. Qpoo - X Loxi)



Answers

From the equilibrium concentrations given, calculate $K_{\mathrm{a}}$ for each of the weak acids and $K_{\mathrm{b}}$ for each of the weak bases.
$$
\begin{array}{l}{\text { (a) } \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} :\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (b) } \mathrm{ClO}^{-} :\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}} \\ {[\mathrm{HClO}]=2.38 \times 10^{-5} \mathrm{M}} \\ {\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (c) } \mathrm{HCO}_{2} \mathrm{H} :\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 M} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} M} \\ {\left[\mathrm{HCO}_{2}-\right]=9.8 \times 10^{-3} \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (d) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} :\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]=0.233 \mathrm{M} ;} \\ {\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}}\end{array}
$$

In this question we have to calculate a for each week assists and KB for each week based from the equilibrium concentration. First one is C A S T C W H. Okay, the equation of for the association of c s d c wh is plus age to war. CS three, plus is trio plus. Okay, the key A is Concentration of C is three CW and concentration of a studio plus And you handed by CSTC 08. Now the concentrations are given, this is 1.34 times 10 to the par minus three and hydro knee um concentration also 1.3 14 to depart minus three Divided by concentration of 60 c double 8, 6 Times 10 to the Power -2. There are four K A. The value is 1.82 times 10 to the power minus five. Okay, this is for the first one. For the 2nd 1 we have to write the equation for the association of C. L minus R. Plus age to war. H. C. L. O. Plus or eight here. K. B for cielo minus concentration of A C. L. O. Concentration of Oh it's my last divided by cielo minus the concentration of a sociology rain. 2.3. 8 times 10 to the bar -5. and concentration of H -4 times came to the bar -4, divided by 0.273. There is artists 3.49 times 10 to the power negative. eight for the third one A C. W. H. The equation for the association or a C. Double eighties plus age to war is your to minus ECU was plus a studio plus ECU was then K. A. Is equal to concentration of 80 to minus times is to your class divided by concentration of a C developed age, Concentration of issue to minus given as 9.8 times 10 to the 4 -3. Nine point they tend to be a negative three for hydro Nehemiah. And for a see wh 0.5-4 there are four key A. Comes 1.83 times 10 to the power negative four. and the last one is for C. Six H five N 83 plus C six H five and H. Two A. Quasi plus is through your plus. Yeah. Okay concentration are C. Six H five N eight to times is to your plus. You heard it by sea seats 85 83 blast this is six. Okay. Now the concentrations are giving therefore G 2.3 times 10 to the 4 -3.3 into the 4 -3 1233. So that gave Hello is 2.27 times 10 to the bar negative fine. Yeah.

In this question uh Florence as it is given with this K. F. Hello? And if concentration we have to find the concentration of is theo plus cielo to minus. And the acid. Eight cielo two at equilibrium. Okay. eight cl or two plus water is still blast Plus. CL.-. Now let's make that ice table. Mhm. Mhm. Yeah Initial concentration is 0.0 to aim judo, judo. Now jeans concentration -1 plus six plus six. And equilibrium concentration +010 to minus six X. X. Let's write the expression for K. A. K. A. Is concentration up is still plus time C. L. +02 miners and divided by the concentration of A C L. O two mm. Now the value of K is given 1.1 times 10 to the -2 eggs X. And this is 0.010 to -X. There are four mhm X squared divided by 0.0 to minus six is 1.1 times 10 to the power negative Now is 02 is a strong acid. So percent organization will be not higher. So will not make any simplifying assumption here we'll use the quadratic formula to solve for X. There are four x square plus 0.211 eggs minus 0.2 equals two. This will be the quality equation. Now we have to solve for X. Now here A is one, B is 0.11 and C is negative 0.0 to two. Now we'll use this formula X. S minus B plus minus route over B squared minus four A. C by two A. If we substitute of hello for a B N C will get X. A week. Was 0.0 103. or negative 0.0213 and will ignore this negative value and will accept the positive value. So there are four concentration of high Dona Mayan is equal to concentration of C. L. A. To minus ion is 0.0103 mm. And You can write it this way 1.23 times tend to depart negative to um and concentration of HCL. is 0.0 to -X, which is 0.0103. So it will be 0.0097. a.m. Or we can write it like this 9.7 times 10 to the power -3. What?

Let's solve for the pH of a mixture of first part of the mixture is acetic acid and so acetic acid. You nice table, but 1000 by its X plus X plus x went zero for 310 My ex ex decks K is equal to H 30 plus C h three c o minus over C H three c o h were given the k a here, which is 1.8 times 10 to the negative five. Zeke with you Ex X. What ones? You're minus X. Let's go ahead and solve this for X on. We'll get a quadratic. I'm gonna put this into my equation solver. 1.8 times 10 to the minus five is equal to x squared in the numerator port 10 minus sex nominator solving for X Here, ignoring the negative roots, we get 0.0 13 and second we gas it Here is Ben's Ilic acid and Zork acid is C six gate five c o plus H +20 H. Three plus at c six. H five c o minus said after nice table here and we're gonna use why to describe the change point ones remind us why, Why and why. Katie. Here is each three old plus c six h five c o minus oversea 65 CEO each We're given the K here 6.5 times 10 in the negative fire in this vehicle to why? Why 0.10 minus y again solving For why here. So the quadratic equation 6.5 times 10 to the negative five and X squared over 0.1 minus X. Ignoring the negative roots, we find that it's equal to 00 to 5 for why so the polarity of each three oo plus is X plus why x we found waas 0.13 plus 0.25 So for X, we find that it is equal to 0.38 and therefore Ph would be negative log 0.38 which would give us a pH of 2.42

Chapter 16 Problem 58 from the book Chemistry. The Central Signs The question states the K a constant for each cielo to is 1.1 times 10 to the negative, too. It access to calculate the concentrations of H 30 plus the Cielo to minus and h Cielo to had equal. They're real when given the initial concentration of Tzar point, sir. 1 to 5 more clarity. So the first thing we have to do is write off. Write our equation for this reaction. So we have HCL 02 It's an Aquarius in addition to water, which then creates C l o two minus plus h +30 plus right in the question acts for all the concentrations at equilibrium. So we know we had to do a nice table were the ice table just stands for initial changed and then the equilibrium So equilibrium you just add the two different things up at the initial on the change. So we first started off with initial concentration a 0.125 hilarity. We didn't have anything to start off with water and water. It does not enter change anything in our table, so I was just cross it out and then we don't have any certain concentrations of our Cielo to O r each for you. Plus, so now we talk about change. So what's actually changing in the reaction to cause our each seal of two to go to equilibrium? So for this to happen, we're actually gonna have tto lose some of our HDL or two since we don't have any starting products and they were going toe at X to both of these sides. In essence, we have that within just add these two rows up, so has their 20.125 plus and minus eggs. And then we just have X and X. Now, since we have the general basis for our equilibrium now we need to write R K expression. So we have K is equal to our products or reactant. And so our products is R C L 02 minus times are each 30 plus. And then our reactant is the H. C. Uh oh, too. And once again, water is not included because it is not a Prius. And so now we just plug in the values we have at the equilibrium So for both Cielo to minus and h the a plus that they're both ex. So I'm just going to write X squared and then for HCL 02 we have 0.125 minus x and in the question they gave us. But our K value is in its 1.1 times 10 to native to and I would just solve for X into solve for X you can use the solver function in your calculator or you consulted Algebraic Lee. But another thing to point out is that you could also there's some really good videos on how to use the solver function of finding accident year on different websites. So what we're going to do is I'm just gonna multiply my ex ever here and dio 1.1 times 10 to the negative too. Is there a point, sir? 125 minus X is he goto X squared and then multiply my oh, these two numbers out together and get 0.7 five and then minus the 1.1 times 10 to the night of two X is equal to X squared. And now I have a quadratic formula, and I would just bring it all to the same side. And so I have X squared plus 1.1 times 10 to the negative two x minus. Is there a point in Sierra +075 is equal to zero and then I would sign is equal to x mean one bee is a good 1.125 and C is equal to this value. And do the staggering I mean the project formula to suffer ex um so our ex actually ends up being seven point for five times 10 to the negative three. And I got this number by Jane the solver function in my calculator. Or like I said, you can also do a hand out by solving for the Pythagorean theorem. We're in this case are a hey is eagle toe one b is equal to 1.1 times 10 to the negative, too. And C is equal to negative. Zahra Point, Caesar's heir of 75 And in just solving for exactly. So now, since we have X now we can solve for our equilibrium. Concentrations down here and since way have excess just values. You're here than R C L 02 and HBO Plus is just 7.4 five times since a negative three. And the sinking for this one and our each Cielo to is going to be their points or one, 25 minus R x value. And I put that into my calculator end. I have the final member of 0.50 for five. So these are my final concentrations at equal there, Bram.


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