suppose-you-mix-1-mol-of-sulfuric-acid-mathrmh_2-mathrmso_4-with-1-mol-of-sodium-hydroxide-naoh-wh-2

Name: suppose you mix 1 mol of sulfuric acid mathrmh_2 mathrmso_4 with 1 mol of sodium hydroxide naoh wh 2
Uploaded: 2020-10-15
Duration: 3 min 12 s
Channel: Numerade
Description: Suppose you mix 1 mol of sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4},$ with 1 mol of sodium hydroxide, NaOH. Why does the $\mathrm{pH}$ of the solution remain below 7 ?

Question

Suppose you mix 1 mol of sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4},$ with 1 mol of sodium hydroxide, NaOH. Why does the $\mathrm{pH}$ of the solution remain below 7 ?

Vishal Sharma · Accepted Answer

so in problem. Five each more off sulfuric acid that is etched two s 04 forms two moles off hydrogen in in solution Two moles off hydrogen irons in solution in solution while each wine each more lof sodium each mall off sodium hydroxide That is any which forms one mole off hydroxide ein forms one mall off hydroxide ein hydroxide irons. So the final solution. So basically this is going to happen like sulfuric acid. So it forms to each plus science and the language. It will form one hydroxide iron in this solution. So this is as a four to minus and this is Annapolis. So do tow this platform. To which place and as as it is, say that each model sulfuric as it formed to Matloff hydrogen nine. That means this I am so this is the H plus iron is thehyperfix login iron and this O H minus iron is the hydroxide ions, hydroxide ions. So therefore we can say that the final solution the final solution the final solution has more hydrogen final solution has more hydrogen irons, more hydrogen irons, then hydroxide I in, then hydroxide irons. So it is acidic. So It is athletic, and the pH is less than seven and the pH is below seven. So this is our final answer. Yeah.

Susan Hallstrom · Answer

if this problem, we&#x27;re going to look at neutralizing mono and di product acids. The question is we have a quick solutions of hydrochloric acid and sulfuric acid, and you can see that hydrochloric acid is a Monta protic Acid and sulfuric acid is a dye product. Gas it into each of these and they&#x27;re at the same Moeller concentration. So they&#x27;re both acids similarity, and we&#x27;re going to try to neutralize these with sodium hydroxide, which is a quits. Whoops. Not that I&#x27;m wrong in a. O. H. A Q and were asked which of these will require mawr sodium hydroxide. Which acid requires mawr of the base and we can see it will be the dye product acid. But we&#x27;re going to prove that now. Okay, for this reaction, my products are a quiz sodium chloride and liquid water. You can see that I&#x27;ve got a 1 to 1 to one ratio 1 to 1 to 1 to 1 ratio. And for my second reaction, my products are a quiz sodium sulphate. Notice that what my formula is right there and liquid water. Now, this is not a balanced chemical equation, and I&#x27;m gonna have to have a two right here and a two right here so you can see that my die product acid requires twice, two times as much any O h for neutralization. And I think that&#x27;s it. Problem solved.

Ronald Prasad · Answer

H two s old three is a weak acid. Being a weak acid will exist in equilibrium. It&#x27;s just so three mine and sulfuric acid is a strong acid which will undergo a complete ionization. And this would be yes, for two minus on. Uh, if we have one Moeller beach in a weak acid, the polarity of the H plus here is going to be much less than the polarity of the H two s +03 So we can say that this age plus here is negligible and we can ignore it. Um, but for each to us. So for the polarity of the H plus, since it&#x27;s a 1 to 2 more ratio that would be equal to two Moeller here. So we cannot ignore that because it does not exist in equilibrium. Ah, and following the definition of a strong acid compared to the definition of a weak acid

Amy Jiang · Answer

Okay, so we have that h two s 04 reacts to produce any too. Plus, it&#x27;s Joel. So we have that don&#x27;t point Field 743 times 0.438 is equal to 0.3 to 6 moves of any. So it&#x27;s more supply that by 1/2 to get the moves of H two s before a 0.1230 most of H 204 and then now calculating its polarity, we have the most of it to us before divided by its volume and military. So we multiply that by 1000 to return it into leaders, which is the 0.3561 polarity.

David Collins · Answer

if we were to titrate a strong acid with a weak base or trade a weak base with strong asset. Either way, at the equivalents point, the weak base is going to be turned into its conjugated acid, which will be a weak acid. Therefore, the equivalents point we have a weak acid and a weak acid produces an acidic solution with a pH less than seven.

Colton Brushwood · Answer

here we are given to Polly product acids or to die product acids. We have sulfuric acid and self, your assassin and we are told that when we calculate the ph of one molar sulfuric acid or really any concentration of sulfuric acid, we cannot ignore the H plus ions that come off by sulfate. And so once we remove our first proton, we have to also calculate the amount of protons that are that that come off of the an eye on HS 04 minus two form eso for two months. So we&#x27;re pretty much told that we have to calculate this I on two as well as these first hydrogen that came. But when we talk about so furious acid, we don&#x27;t have to calculate this proton that comes off of HS of three minus. So we only have to calculate the first proton that comes off of H two s 03 not the second. And we want to know why that ISS So, in order to know why we can ignore the proton that comes off of the by sulfide and on first we need to look at their initial assets. So we have sulfuric acid on the left and sulfurous acid on the right on the left. So if uric acid has a K A of 1000, so that&#x27;s one times 10 to the third. We rarely see acids. At least that we work with that are this strong. Sulfuric acid is very strong. Okay, of 1000 pretty much means no matter what it&#x27;s gonna want to get rid of its protons. The base is just that stable. When we look at sulfurous acid, the K is still pretty high. It&#x27;s 1.3 tons, 10 to the negative to, but it&#x27;s not quite as high as 1000. It&#x27;s still much, much weaker than sulfuric acid. So that&#x27;s our first proton. So that is for you know, this very first proton, this proton in this proton to remove those on both of these is easy. But when we go down and look at our by sulfate and I on and by soul fight and are by sulfate, and I on now has a K of one times 10 to the negative, too. So as we know from previous working with acids, that number is still pretty high for a K A value, especially when we talk about something that&#x27;s already lost. A proton for a Dia product acid that is still a very high K a value now buy soul fight. It&#x27;s K of its second proton. So we&#x27;re talking about this second proton on both of these, the second one for By Soul Fight it&#x27;s K is 6.3 times 10 to the negative eighth. That&#x27;s a pretty small number that pretty much means it&#x27;s not going to want to let go of that brutal. In fact, it may even reform a little bit with its first proton. It&#x27;s a weaker acid. So really, this is what makes that apparent the k a values of by sulfate and by soul fight. And as we know in previous examples, if we&#x27;re working with ice tables, so we&#x27;re doing well, I guess ice or rice tables. So if we do rice slash ice tables, we have our what we call our approximation. If I could spell approximation right, that&#x27;s approximately right, but not quite so. We&#x27;ve got an approximation rule for our ice tables, and that means remember, when we are setting up our K expression, or KB, we have our H 30 plus plus our A minus over our age. A Valium. Remember, for this expression that a J is pretty much our acid, minus the amount that dissociates so that X value that we typically would approximate four down here in the denominator because of this small K a value we can&#x27;t do that with this 10 to the negative to on sulfuric acid. So for our second Proton were not able to approximate this X out of the bottom of the equation. That X makes a difference. The amount of protons that disassociate from the by sulfate and I on it&#x27;s still a large amount. It is still affect the pH in a pretty sizable and measurable way. It&#x27;s not going to go unnoticed if we leave this X out of RK expression for sulfuric acid. So we&#x27;re trying to find the pH of sulfuric acid. Not only we don&#x27;t have to do a nice table for the first proton. We know the first proton disappears, but we do have to do a second ice table in K expression for the by sulfate and I on, because it will still lose enough protons to alter the Ph in a way that we could measure easily. So that&#x27;s the explanation for why we can&#x27;t ignore the proton that comes off of by sulfate. But we can ignore the one that comes off by Isil fight. It all goes back to our approximation rule that we use when we&#x27;re doing K expressions four acids. And that means RK is a very small number for by soul fight, but still a large number for by sulfate. So we do still leave that X in the equation. We do have to do an ice table for the by sulfate and I on not by so fight.

Problem

Suppose you mix 1 mol of sodium hydroxide, NaOH, …

Problem 5 Easy Difficulty

Suppose you mix 1 mol of sulfuric acid, $\mathrm{H}_{2} \mathrm{SO}_{4},$ with 1 mol of sodium hydroxide, NaOH. Why does the $\mathrm{pH}$ of the solution remain below 7 ?

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