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(This thc solution? H Submit Answer OWLV2 Rcfercuces' } elass nnbat cololectodcalan Tables 1 and pH Amother anSQGr Ipmerhetered Version 0 V bccame 1 widely Int...

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(This thc solution? H Submit Answer OWLV2 Rcfercuces' } elass nnbat cololectodcalan Tables 1 and pH Amother anSQGr Ipmerhetered Version 0 V bccame 1 widely Into Fonmea 1 solula which 1 8 [ 8 [ 1 base fortu VL ofthe and bast wbat [ 1 oftia

(This thc solution? H Submit Answer OWLV2 Rcfercuces' } elass nnbat cololectodcalan Tables 1 and pH Amother anSQGr Ipmerhetered Version 0 V bccame 1 widely Into Fonmea 1 solula which 1 8 [ 8 [ 1 base fortu VL ofthe and bast wbat [ 1 oftia



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Chose the correct code \begin{tabular}{|l|l|l|c|} \hline \multicolumn{4}{|c|} { Column-I } & \multicolumn{2}{|c|} { Column-II } \\ \hline (P) & $p K_{b}$ of $X^{-}\left(K_{a}\right.$ of $\left.\mathrm{HX}=10^{-6}\right)$ & (1) & $6.9$ \\ \hline (Q) & $p K_{b}$ of $10^{-8} \mathrm{M}$ HCl & (2) & 8 \\ \hline (R) & $\mathrm{pH}$ of $10^{-2} M$ acetic solution (Take $K_{a}$ of acetic acid $\left.=1.6 \times 10^{-5}\right)$ & (3) & $10.7$ \\ \hline (S) & pOH of a solution obtained by mixing equal volumes of solution with pH 3 and $5 .$ & (4) & $3.4$ \\ \hline \end{tabular} Codes: $\mathrm{P} \quad \mathrm{Q}$ $\begin{array}{ll}\mathrm{R} & \mathrm{S}\end{array}$ $\begin{array}{lllll}\text { (a) } & 1 & 2 & 4 & 3\end{array}$ (b) $4 \quad 3$ $2 \quad 1$ $\begin{array}{lllll}\text { (c) } & 2 & 1 & 4 & 3\end{array}$ $\begin{array}{lll}\text { (d) } & 1 & 2\end{array}$ $3 \quad 4$

Hello and welcome to this video solution of numerous here. If you have H tribute for as a try protic acid the week a poetic has it right. And you have to calculate the approximate ph of 0.1 molar of NH two H p 04 and is calculated using the expression. So we we know that it's option B rate which is the average of Mhm A two Plus PK it three, right. PK two refers to the dissociation of in a H to P 04, and PK three refers to the decision of And at two HB 04. Right. Uh so we select option B. I hope scared to you and have a very good rest of the day.

So here are going to be calculating Mill Aridjis Um, given the mill Aren t and K a of other solutions. So for a V of age Cirio plus there's a strong acid. So it's 30 physical stage five, which is a equal to 0.0 nine to allergy. Um, we have the reaction. We have H C three h 502 plus h +20 those two c three h 50 to minus plus h Theo Plus. So this is your ex and we know this is your 0.892 plus X and this is your a 0.275 minus maximise like so, the K A here is equal to X time zero boy zero a nine to trust specs divided by 0.275 mining specs, which is equal to about 0.92 x. Is there a point your 892 by by 0.275 which means that X is equal to four times 10 to the negative fit polarity, Um, and the hilarity of aged 30 plus Z equals 0.892 So now we can use this and calculated for the rest of them. So be so h minus, which is equal. Okay, w over where did that, which is equal to 10 to the negative 14 over. Was there a 0.892 which is equal to 1.1 times tender than I get 13 polarity. Never see that we know is equal to X, which is equal to you four times tended and I get 1/5 polarity. And lastly for D you have I minus is evil too. H i, which is equal to zero quay 0892 polarity.

Similar to the previous problem. We need to identify what is stronger for part A. We have potassium hydroxide and we have barium hydroxide. They are both strong bases, so they contribute equally by completely dissociating according to their concentrations. Potassium hydroxide is 0.5 Moeller barium hydroxide is 0.15 Moeller, but we get to hydroxide is for every barium hydroxide. So the total hydroxide concentration from both of these strong bases completely dissociating is going to be equal to the contribution from potassium hydroxide at 0.50 Moeller and two times the contribution from barium hydroxide. Because there are two hydroxide for everyone. Barium hydroxide. So two times 20.15 Mueller and we get a total hydroxide concentration of 0.80 Moeller. So the pH is going to be the negative log of the hydro knee, um, concentration, which is going to be the hydroxide concentration we just calculated divided into K W. We take the negative log of that ratio and we get a pH of 12.90 for the next one. We recognize that we have ammonium, which is a weak acid, and we have hydro satanic acid, which is a weak acid. So we have to weak acids, and we want to figure out which weak acid is stronger. So we'll solve for the que a value of ammonium by taking the K B value of ammonia, which we looked up and dividing it into K. W. And we get 5.68 times 10 the negative 10. Then we look up the K a value for hydro satanic acid, and it's essentially the same. 4.9 times tend to the negative 10. So one way to handle this is to assume that both of them are behaving to the same extent. And we can some there to concentrations together to get a total weak acid concentration and then just use an average K value something around five times 10 to the negative 10. Then the hydro knee. Um, concentration would be equal to some average K, a value multiplied by the some concentration of both of the weak acids with essentially equivalent K A values. And we get a hydro knee, um, concentration of about 1.4 times 10 to the negative five, and then we take the negative log of that and get a pH of 4.85 for the next one. We have a solution that contains rubidium hydroxide, which is a strong base, and we have sodium bicarbonate where bicarbonate is AMFA, Terek and can behave is an acid or a base. So it is likely that if we have the strong base rubidium hydroxide in a solution with potassium carbonate that that rubidium hydroxide is going to react with the bicarbonate and produce just carbonate. However, the solution in the solutions manual does not consider that complex reaction. So I'm not going to either. I'm just going to assume that the solution has already reached equilibrium, and these are the concentrations of rubidium hydroxide and sodium carbonate after reaction has occurred. If that's the case, then we focus on the stronger of the two, which is rubidium hydroxide, and we get a hydroxide concentration equal to the rubidium hydroxide concentration of 0.75 So the pH will be equal to the negative log of the hydro knee, um, concentration, which is the hydroxide concentration coming from rubidium hydroxide divided into K. W. And we get 12.88 then for D. We see that we have a solution of the strong acid perk Lorik acid and a solution of the strong base potassium hydroxide. Now, in this case, in order to get the correct answer, we do need to recognize that there is a reaction between these two and that some of the potassium hydroxide is going to consume well, all of the potassium hydroxide is going to consume some of the perk Lorik acid. So the concentration than of HCL 04 is going to be the 40.88 minus what reacts with potassium hydroxide. 0.0 to 2 and we get a concentration then of the strong acid, um, folkloric asset of 0.66 So the ph, then it's going to be the negative log of 0.66 or 1.18 Then for the last one, we see that we have a solution of sodium. Ah, hypochlorite hypochlorite came from the acid Heiple Cloris acid. So itself is a base, and then potassium iodide does not do anything to affect the pH of the solution. Potassium coming from the strong based potassium hydroxide and iodide coming from the strong acid hydro hydro I otik acid. So we focus on the c l o minus to determine the pH of the solution. We can calculate the KB of Cielo minus by dividing DK a value of HIPAA Cloris acid into K W to get a K value of 3.45 times 10 the negative seven. Well, then get our hydroxide concentration by taking the K B value multiplied by the concentration of Hippo. Cloris, I'm sorry. Multiplied by the concentration of hypochlorite ceelo minus at 0.115 Take the square root of that and we get 1.99 times 10 to negative four. This is definitely less than 5% of the 50.115 So we don't need the quadratic we can then get the pH by taking the negative log of the hydro knee. Um, concentration, which is the hydroxide concentration we just determined divided into K W. And we get 10.30

Problem. 38 At 25 divisions years. We know that concentration of hydrogen iron Multiplied by concentration of hydroxy and equals two. One into 10. to the power -14 and ph equals two. Sorry ph plus The. US. equals 2 14. Now use this lesson ship for solving table. And we know that if beards less than seven then solution is pathetic. And if Viet greater than seven solution is basic. No this data. This data is given enough questions. So first we use this formula and we in this poem where we put the sorry sorry this formula we put the value of peers and we get the value of P watch that is eight point seven. Hi. No. We know that ph equals two minus log as and concentration. So we can take minus log of this quantity. And we We get the value of 109 consent is um that is 5.6 into 10. To the power -6 M. Now this value is put in this formula here and we get the concentration of words And where's concentration is 1.8 into 10. To the power minus nine a.m. We know that. Sorry. We clearly see here the ph of solution is less than seven. So solution is I said it. No. In the second part Helio p O H is given. So we simply use this formula again and put the value of us and we get the value of ph that is 11 point 98 Now we can take minus sorry, We can know that had an and concentration wherever equals two minus. Sorry, sorry P. S. Equals two minus log as and concentration. So we can take minus log of these periods and we get the concentration of hydrogen. So opera calculating we get the value of S. And concentration concentration is 1.1 into 10. To the power minus 12, 12 AM. Now this value of pattern and concentration we put here in this question and we get the value of hydroxy and concentration that is 9.6 in 210 to the Power -3 M. And we clearly see here Ps is greater than seven. So solution is basic, no In part 3rd concentration of production and is given. So we simply use this formula and put the value pattern and in this equation and we get the concentration of hydroxide iron, that is two point three into 10 to the power -5 them. Now we can take the minus log of harder than iron, so minus log of this quantity minus log of 4.4 inches under the power managed. And um and we get the value of ph that is nine point 36. Now we use this formula and put the value of periods and we get the value of the words, that is four six ft. Now we clearly see here Ps absolution is greater than seven. So solution is basic in nature, no in part last part, concentration of ways was negative and is given. So we use this formula and put the value of us and and we get the Concentration of partisan iron. After putting the value, we get the value of partisan and that is one two in 10 to the power minus 13. No, we can take the minus log of traditional and concentration you get and we get the value of period that is 12.93. Now this value put in this formula and we get the value of pos that is After calculating 1.07. Now we clearly see here ph absolution is greater than seven, that is 12.97 So nature of solution is basically, and this is our final answer for this question.


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