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(51) Problem Set # 14 (Ch 14)Exercise 14.47Four solutlons ol unknown HCl concentration are Iitrated wlth solutlons of NaOH. The lollwing tablo lists the volumo ol e...

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(51) Problem Set # 14 (Ch 14)Exercise 14.47Four solutlons ol unknown HCl concentration are Iitrated wlth solutlons of NaOH. The lollwing tablo lists the volumo ol each unknovn HCl sclution, Ihe volume ol NaOH solulion required t0 reach Ihe equlvalence point, and the concentralion of each NaOH solution.HCI Volume (mL)NaOH Vclume (mL)NaOH] (M) 0.1281 M 0.0932 0.1148 0.1385 M24.00_mL 14.00 _L 25.00 mL 3.00 mL33.,44 mL 25.22 mL 12.88ML 7.88 mL

(51) Problem Set # 14 (Ch 14) Exercise 14.47 Four solutlons ol unknown HCl concentration are Iitrated wlth solutlons of NaOH. The lollwing tablo lists the volumo ol each unknovn HCl sclution, Ihe volume ol NaOH solulion required t0 reach Ihe equlvalence point, and the concentralion of each NaOH solution. HCI Volume (mL) NaOH Vclume (mL) NaOH] (M) 0.1281 M 0.0932 0.1148 0.1385 M 24.00_mL 14.00 _L 25.00 mL 3.00 mL 33.,44 mL 25.22 mL 12.88ML 7.88 mL



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Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown $\mathrm{NaOH}$ solution, the volume of $\mathrm{HCl}$ solution required to reach the equivalence point, and the concentration of each HCl solution. Calculate the concentration (in $\mathrm{M}$ ) of the unknown $\mathrm{NaOH}$ solution in each case. $$\begin{array}{lcc}\hline \text { NaOH Volume (mL) } & \text { HCI Volume (mL) } & \text { [HCI] (M) } \\ \hline \text { (a) } 5.00 \mathrm{mL} & 9.77 \mathrm{mL} & 0.1599 \mathrm{M} \\\text { (b) } 15.00 \mathrm{mL} & 11.34 \mathrm{mL} & 0.1311 \mathrm{M} \\\text { (c) } 10.00 \mathrm{mL} & 10.55 \mathrm{mL} & 0.0889 \mathrm{M} \\\text { (d) } 30.00 \mathrm{mL} & 36.18 \mathrm{mL} & 0.1021 \mathrm{M} \\\hline\end{array}$$

To answer this question. We need to know what the stock geometry of the Thai trey shin reaction is. We have HCL reacting with sodium hydroxide to produce sodium chloride and water. This is a one toe one reaction where one mole of HCL reacts with one mole of any O. H. To calculate the concentration of HCL that is required, um, or that is found in the volume of HCL that is provided. We need to know what the definition of concentration is. It will simply be moles of HCL, divided by leaders of HCL solution. So in the Thai Trey Shin, we have 28.44 mL of sodium hydroxide will convert that those mill leaders two leaders, so we can then use the concentration of sodium hydroxide to convert our leaders to moles. This will tell us the moles of sodium hydroxide that reacted with the HCL that was present. Once we know the moles of sodium hydroxide because the reaction is one toe one, we can then calculate the moles of HCL that reacted. So we have just determined this numerator value right here molds of hcl. We then need to divide that by the volume of the HCL solution that was provided at 25.0 mL. But it needs to be in leaders, so we'll divide that by 1000 in order to get leaders. So when calculating the concentration of HCL is going to be the moles of HCL divided by the leaders of solution, this calculation gives us moles HCL, starting with the volume of N a. O. H that reacted with the HCL with the event. Then divide that by the volume of HCL that was provided. This gives us a concentration of HCL 0.1400 Mueller. We'll do the same thing with the other calculations. We'll start with the volume of sodium hydroxide that reacted with the HCL in milliliters, converted to leaders by dividing by 1000. Then convert the leaders of any O H two moles of any ohh by multiplying by the molar ity of the solution and then with the moles of any O. H. We can calculate the moles of HCL because we know the reaction has a 1 to 1 stoke eom A tree is explained before that will give us moles of hcl. We then divide that by the volume of HCL Solution, which is 15 million leaders that we need to convert to leaders by dividing by 1000. And we get a concentration of HCL of 0.1375 Mueller for the next one. We have 14.88 mL of any Ohh that reacted with our HCL solution will convert those mill leaders into leaders by dividing by 1000 and then convert the leaders of the O H solution into moles of any ohh by multiplying by the molar ity of any Ohh and then last of all, convert the moles any O h two moles hcl based upon the story geometry of the reaction being one toe one Now that we have moles, HCL will divide that by the volume of HCL solution to get more clarity. The volume, however, needs to be in units of leaders. So we'll convert our mill leaders into leaders by dividing by 1000 and we'll get a polarity of 0.876 for the last one. We have 6.88 mil Leaders of any O h that reacted with our HCL will convert that to leaders by dividing by 1000 and then convert the leaders of the O H solution two moles of any O. H by multiplying by the polarity of any O. H. Then when we have moles, any ohh will convert it to moles HCL with historic geometry of the reaction one toe one and then we'll divide that by the volume of the H. I'm sorry, the HCL solution, which was five mil leaders and we'll convert that to mill leaders. I'm sorry the mill leaders still leaders by dividing by 1000. This will then give us moles hcl per leader of the HCL solution or a molar ity of 0.18 to 3.

So in this example, we're looking at a mixture of HCL and NH where we have performed tight rations where we're using our title ation valleys Andi volumes to determine the concentrations so we can look at a concentration of sodium hydroxide fast. So this the concentration of sodium hydroxide. We can start off with the moles, which is two times theme moles of H two s 04 So we have 7.82 million moles. We can then combat this two volume which is 18.4 mL. Then we have a concentration which is moles divided five volume. So we take 7.82 divided by 18.4 gives us not 0.4 to 5 moles. We can then move on to the concentration of HCL. So we look at the moles off sodium hydroxide and the moles of HCL, which is concentration multiplied by volume which is 11.69 Minimal volume of HCL in this example is 100 mL now, So we take the malls divided by the volume to generate a concentration of not 1000.1168 moles

Hi everyone so far in this question that's that is you have given the solution of selling an image and must be to remind their concentration. Youth 27.1 ml orphanage to titrate 100 develop it sell and 18.4 ml of an image. To detract 50 ml Off 0.0782 More arrest yourself or find the unknown concentration. So what they're given. So first of all will rate given data volume of an H. That is equal to 27.5 ml. That is equal 0.0 to some five leader. That is divided by thousands. Then volume up at seal. That is equal to under animal. And that is equality 0.00 day to the next to a limo. For an average That is equal to 18.4 a mill. Come on. That is equal to 0.0184 leader them my lima vegetables self for I was so important 96 quantum repayment 0.050 L. Then consolation of vegetables self. That is equal to 0.0782 Moeller. So you have to find out concentration of any ridge. I was 14 concentration of an image question Emma. Then concentration of Hcl question mark saw a part. I love it concentration of any wage of the neck of age so first will decline and now finally right and that is equal to okay uh before that train. Nfs to yourself. Fourth enough that's two SA. For that is equal to concentration into volume. It's a concentration 0.7 82. Volume 0.50 And that is equal to zero point $00391 then and of an average second. And of any ohh that is equal to do into normality of H. two a cell phone And that is equal to two into 0.00391. And that is equal to 0.00782 malls. Yeah. Republic most. I love it. Therefore concept her points concentration of N. H. That is equal to and upon will that is equal 0.7 82. Divided by volume 0.1 84. And that is equal to 2.4 to $5. And therefore the property market somewhere consolation of any wage. All right Consideration of an average 90 is equal to 2425 mothers then be part do part concentration of itself. But Republican talking concentration of its seal. That is equal to end of any wage. Thanks and the rich that is equal to make me and of excel. And that is equal to consolation into volume. That is equal to let me help you know .4 to fire you In 2 0.02 17 for you. That is equal to I see no point 011688. More. Yeah, mourns thanks. Mhm. Okay, Mom's then concentration of that seal also. Okay. Mhm. Therefore, concentration of At cl that is equal to any upon way and that is equal to 0.011688 divided by we .1. What is the problem? And that is equal to 0.11688 Mauler. And therefore therefore concentration of and seal 19 sickle. 0.1168 eight moller. Thanks.

This is question 40 from chapter 15 and neutralization problem, and we're trying final polarity. Uh, the base in this case, sodium hydroxide. Now the neutralization equation is in a V A is equal to and be DP A scenes were acid and B stands for face, so they give us both and and be for the acid, which is 50 no leaders times the Moeller 0.104 and his hydrochloric acid. That's what the acid is with this column and that is gonna be equal 2 48 0.7. No leaders of the base and want to put X for the military base since we don't know what okay, when you want to play 50 and 500.104 favor based, the most of this is going to be five point to No, actually, that now we can divide by the volume of the base to find the mole ari of the base. So 5.2 divided by 50 48.7 This is going to come out to be zero 0.1 oh seven most of in a o. H. So that is the polarity of


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