A chemical that is effective in preserving foods with a low ph...

Question

A chemical that is effective in preserving foods with a low ph such as bread is

Accepted Answer

Hello. Everyone in this video, we're being asked if 750 g milligrams of C nine H 10 02 is dissolved in 500 millions of water. Let's go ahead and calculate for the ph of the solution and it's percent ionization. So the reaction that occurs here or what's going on here is that we have the C nine H 10 02. The racks with H 20. And it's an equilibrium with our H three L plus as well as our C nine H +902 minus. All right. So let's first calculate the initial concentrations. So starting off with some information that we should know or that we can find is the molar mass Of the C9 H 1002, which is equal to 150.17 g per moles. Now let's go ahead and sell for more clarity, which is the concentration. So clarity is equal to most over leaders. So we can go ahead and first start off with the moles. So we'll do this in red. So again, we're solving for the moles of C nine H 10 02. So we're given again the 700 million kg of the C9 H 1002. We can go from milligrams into g with a direct conversion. So for every 1000 mg we have one g. Now must find this by the molar mass which will give us from grams to most for every one mole Over C nine H 1002. We have 150.17 g. So now for unit cancelation we can see that the milligrams will cancel and grams will cancel leaving us with the moles of C nine H 10 02 to equal to 4.9943 times 10 to the negative three moles. If I calculating for the molar itty Starting off with the moles here. So 4.9943 times 10 to the -3 moles. We're gonna go ahead and divide this By the millions of solutions that were given. So that's 500 ml. We know that units of modularity contains leaders so we can go ahead and continue this dimensional analysis. So for every 1000 ml we have one leader. We see here that the millions will cancel. Alright then get the molar itty here to be equal to 0.100 molars. Let's go ahead and do an ice table. So of course we have the equation that's going on to the chemical reaction C. Nine H 10 02 which is acquis reacting with liquid H 20. And this is in equilibrium with H 30. Plus And c. nine H 902 -. Alright. Again we have I which is initial c. for change and e. for equilibrium. So for initial concentration here we have 0.0100 molars for Hbo's liquid. So we'll go ahead and neglect all these values for the ice table from our products that are both zero initially. Alright. So for now from a change there are also unknown for maturing material side will always have minus sign. And then from a parasite will always have a plus sign. So since it's unknown, gonna go ahead and use an X. As our value for now. So I have minus X. Here for my side materials and plus X. And plus X. For my two products. So now E for equilibrium is just basic combining these two. So for my first one here that's 0.100 molars minus X. From my right side we have X. And X. All right. So let's actually scroll down again for more space. So we know that our K. A. Value is equal to the concentration of H. 30. Plus multiplied by the concentration of C. Nine H. 902 minus. Go ahead, divide this with the concentration of C. Nine H. 10 02. So, we know the value of K. A. This is 2.2 times 10 to the negative five and just plugging in everything else. So H. Three plus. That's X. This an ion that is also X. And then for my C. Nine H. 10 02 that's 0.100 minus X. Let's go ahead and do some simplifications. So that's X squared. Since we have X times X. The denominator will state as the same. All right. And now we're solving for R. X squared here or for our X. So we'll have 2.2 times 10 to the negative five times are 0.0100 -1. We'll see here that we just multiply both sides by the denominator of our fraction. So this equals two x squared. Then we go and distribute this on the left side of the equation signed to give us 2.2 times 10 to the negative seven minus 2.2 times 10 to the negative five X. Which equals then two X squared. So we get then zero equals two X squared plus 2.2 times 10 to the negative five X minus 2.2 times 10 to the negative seven. So we can see that we have a quadratic equation here. So we go ahead and use the quadratic formula. We'll go ahead and do this in blue on the right side. So then we get that X. Is equal to the negative three negative B plus or minus the square root of B squared minus four A. C. And it's all divided by 2 8. Alright, so now plugging in my values, we have negative 2.2 times 10 to the negative five plus or minus the square root of 2.2 times 10 to the negative five. And this is squared -4 times one Times -2.2 times 10 to the -7. It's going to be divided by two times A. Which is just one. Alright so again we're just doing a bunch of simplification. So we see here that will actually get negative 2.2 times 10 to negative five minus 9.3834 times 10 to the negative four divided by two. Get the X. is equal to negative 4.8 zero 17 times 10 to the -4. Alright and then this then is equal to well she actually have to divide both sides by this negative one. So we should have actually gone 4.8017 times 10 to the negative four. So since we have this equaling to the concentration of my H. 30 plus, so just this portion right here but you actually use this to sell for a ph value. So for my equation for P. H. Is equal to the negative log of the concentration of H. 30 plus. Which we have will simply put that into the equation or the formula rather. So we're taking the negative log of 4.581 times 10 to the negative four molars. So we can finally get that my P. H. Is equal to 3.34. So this is one of my answers for this problem not cackling for the percent ionization produce and bread. Alright, so again we're starting four ionization. So this equals to the concentration of H. 30 plus Divided by the concentration of c. nine H 1002 As initial state times 100. So we'll go ahead and plug in the values here we get 4.5817 times 10 to the -4, Divide by 0.0100. It's all multiplied by 100. So once you put everything into calculator, we see that the present Finalization is equal to 4.6%. So this is my second and final answer for this problem.

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A chemical that is effective in preserving foods with a low ph such as bread is

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