Consider the unbalanced equation:
(d) What pH is required for t...

Question

Consider the unbalanced equation:  
(d) What pH is required for the reaction to be at equilibrium at 25°C when [I-] = 0.10M and [IO3-] = 0.50 M?

Accepted Answer

Welcome back, everyone. We need to balance the following reaction where di chrome it reacts with one mole of bromide to form our chromium three plus catalon and one mole of liquid bromine and acidic conditions. Calculate the ph of the reaction at equilibrium at 25 Celsius. If our concentration of dye chrome eight equal to our concentration of bromide, both equal 1.50 molar. And our concentration of chromium Catalan is 5.4 times 10 to the negative six power moller. So notice that our equation given in the prompt where di chrome eight cr minus reacts with bromine br minus to form chromium cr three plus. And bromine liquid is an unbalanced redox reaction. And so to balance this out, we're going to need to make two half reactions are first half reaction will be for di chrome eight cr 2072 minus Which forms our chromium three plus catalon. We'll move this below so that we have enough room. So this is our first half reaction and we need to begin by balancing out our atoms. Notice that on the reactant side, we have two moles of chromium were on the product side, we have one mole of chromium. We're going to place a coefficient of two in front of our chromium three plus Catalan on the product side. And also notice that on our reactant side, we have seven moles of water recall that we bounce oxygen using water. And so therefore, we're going to add a total of seven moles of water on our product side. Next, notice that because we added seven moles of water to the product side, we have now introduced 14 moles of protons to the product side. We have none on the reactant side. So we're going to expand our reactant side and add moles of protons to the reactant side. Now with our atoms balanced on both sides of our equation, we now need to balance net charge. Notice that on the product side, we have a net charge of plus six. Notice that on the reactant side, we have a net charge of positive 14 -2, meaning we have a net charge of plus 12. So in order to Balance out our charge, we're going to need to add electrons to the side that's more positively charged. And since we have a net charge of plus 12 on our reactant side, we're going to add six electrons to our reactant side. So we expand our react inside and add six electrons which will now give us a net charge of plus six on both sides of our reaction, meaning our charges are balanced. This completes bouncing our first half reaction. Now we're going to move on to our second half reaction, which is for our bromide, an ion which forms remain liquid as a product. So notice that we have two moles of roaming on the product side, we're going to place a coefficient of two in front of our bromide an ion on the reactant side. And now we have a net charge of -2 on our reactant side. So we're going to need to Again add electrons to the more positive side, which would be our neutral side of our reaction being our product side. And we're gonna get a net charge of -2 by adding two electrons to our product side. This bounces out the atoms in charge in our second half reaction. And now we need to come up with an overall reaction by adding up our two half reactions. So we're going to need to make sure the electrons are balanced. We're going to take the entire second equation and multiply it by a factor of three to give us this new second half reaction where we have six moles of bromide, Which forms three moles of remain liquid and plus six electrons. So now adding up are two half reactions, we come up with our overall reaction where we can cancel out the six electrons from our first half reaction on the react inside with the six electrons from our second half reaction on the product side. And now we're going to bring down everything else that we're left with. And we come up with an overall reaction equal to die chrome eight cr + minus Plus 14 moles of protons Plus six moles of bromide. An ion yields two molds of chromium, three plus Catalan Plus seven moles of water And then plus three moles of grooming liquid. Notice that in our prompt were given a table with the Gibbs free energy formation for each of our re agents. And so we're going to need to calculate Gibbs free energy change for this overall reaction. Recall that we can calculate it by taking the Gibbs free energy change formation for our products subtracted from the Gibbs free energy change formation of our reactant. So beginning with our products, we'll say that delta G and sorry. Just so it's visible. Delta G is equal to the sum of Gibbs free energy change of our products. First, where we begin with our first product being our two moles of chromium three plus multiplied by the free energy formation of chromium. Three plus given in the chart above as negative 143.5 kg jewels Permal. This is then added to our second reactant in which we have seven moles. Sorry, seven, our second product in which we have seven moles of water multiplied by the free energy change formation for water. As negative 2 37.1 killer jules Permal. Then adding to this, we have three moles of our romaine liquid. And actually recall that in our free energy change formation, we only use our acquis re agents only. So this will actually complete the sum of our free energy formation for our products. So now subtracting from the free energy formation for our reactant will begin with our first reactant, which is one mole of die crow mate multiplied by die crewmates free energy formation of negative 1301. kg per mole. Adding to this, we have our second reactant, 14 moles of protons multiplied by the free energy change for a proton given in the chart as zero joules per mole and then continuing below here we have plus our third reactant given in our overall equation where we have six moles of our bromide an ion multiplied by its free energy formation as negative point oh killing joules per mole. Given in our chart above this completes the sum of free energy formation of our reactant. Now we're going to simplify so that we have delta G equal to for the some free energy formation of our product, we should come up with a value of negative and we don't need brackets anymore. We come up with a value of negative 946.7 and noticed that our units left our killer jewels because moles will cancel out. This is subtracted from the sum of free energy formation of our reactant, which results in a value of negative 1925.1 kg because our units of moles will still cancel out. So now from this difference, we're going to come up with our free energy change of our overall reaction equal to a value of negative 21.6 kg rules and this is per mole of our reaction. Now, we're going to recall our second formula for Gibbs free energy change in which we can calculate it by taking negative one times R gas constant R times R temperature in Kelvin times the natural log of our equilibrium constant Casey. Because we need to find the Ph recall that we can isolate this formula for our equilibrium constant by taking both sides of our equation and dividing by negative R T. This will allow us to say that our natural log of K C is equal to delta G divided by negative R T. So plugging in our values beginning with our numerator, we just found our gifts free energy change as negative 21. kg joules per mole. In our denominator, we have negative one times R gas constant R from our textbooks, which we should recall us the value 8.314 jewels divided by moles times Kelvin multiplied by our temperature given in the prompt as 25 Celsius in which we want to cancel out our units of Kelvin. So we're going to add to 73.15 Kelvin for sorry, just to 73.15, which gives us our Kelvin temperature as to 98.15 Kelvin. So we'll plug that down below in our denominator. And so now recognize that in our numerator, we have units of kilograms per mole. But in our denominator, our guests constant are utilizes units of joules per mole. Kelvin. So we're going to need to multiply our numerator by a conversion factor to go from kilo joules in the denominator two jewels in the numerator. Recall that our prefix kilo tells us that we have 10 to the third power of our base unit jewel, canceling out killer jewels. We can also cancel out moles with moles in the denominator jewels with jewels in the denominator. And we're left with Kelvin that we can cancel out leaving us with no units for our equilibrium constant, which is what we want because recall that it should have no units. And so this quotient is going to simplify to a value equal to 8.71383. Now, we just need to get rid of our Ln term and recall that we can use Euler number two, make both sides of our equation and exponents So that it cancels out the Ln term and we will have our equilibrium constant KC isolated so that we have it equal to a value of 6,086.51. Now, we want to recall that our equilibrium constant is therefore interpreted as our concentration of our products divided by our concentration of our reactant. And this is Aquarius re agents only. So we can interpret this based on our overall reaction where K C which we just found the value as 6086.51 equal to our concentration of our products. In which from our overall reaction, we have chromium three plus as our first product raised to a power of two, since we had two moles as the coefficient multiplied by our concentration of our second product, which is our not grooming but rather water and actually not even water because water is a liquid too. And again, we only use what's Aquarius. So this will complete our numerator, which is our concentration of our products in our denominator. We have our concentration of our acquis reactions from our overall reaction in which for di chrome eight, we have one mole of di chrome eight. So we have cr 2072 minus concentration and then multiplied by the concentration of our second acquis react in our bromide, an ion which is raised to a power of six. Since we had six moles of that, an ion on the reactant side multiplied by our third reactant, which is our concentration of H plus raised to a power of since we had 14 moles of H plus in our overall reaction. So note that we're given our concentration of chromium three plus equal to from our prompt 5.4 times 10 to the negative six. Power Moeller were also given our concentration of dye chrome eight equal to our concentration of bromide which both equal the value 1.50 moller as given from our prompt. And we can plug in those values accordingly. In our case, the equation so that we have our K C 6086. equal to in our numerator. For a concentration of chromium three plus, we have 5.4 times 10 to the negative six power molar squared. In our denominator, we have fur die chrome 81.50 Moeller multiplied by our concentration of bromide. 1.50 moler two raised to a power of two or sorry, raised to a power of six in our denominator multiplied by a concentration of H plus which we do not know. And that is what is going to help us get to our Ph. And again, this concentration of H plus is raised to a power of 14. So we need to isolate for our concentration of H plus. So we're going to need to multiply by our concentration of H plus to the 14th power on both sides so that it cancels out on the right what we'll have is R K C 6086.51 times our concentration of H plus to the 14th power equal to 5.4 times 10 to the negative six power molar squared in our numerator divided by our denominator, 1.50 moller times 1.50 moller. Now to isolate for H plus, we're going to divide both sides by 6086.51 so that it cancels out on the left. And what we'll find is that our concentration of hydro ni um or sorry, a concentration of protons to the 14th power is equal to in our numerator, we still have 5.4 times 10 to the negative six power molar squared divided by 1.50 moller times 1.50 Moeller to the sixth power and sorry, that was missing from the previous step, which is also multiplied by 6086.51. So simplifying this quotient will find that our concentration of protons to the 14th power is now equal to the value of our right hand side, which is going to result in 2. times 10 to the negative 12 power moller. And we need to get rid of the 14 power term. And I'd like to actually make a correction. So this product of our quotient or result of our quotient should equal the value 2.80402 times to the negative 16th power moller and then to get rid of the 14th power on the left hand side, we're going to take both sides to a power of 1/14. So that will cancel out the 14 exponents on the right hand, on the left hand side, which will completely isolate our concentration of protons equal to. Now the value 0. moller. Now we can finally get to our ph by recalling that our Ph is equal to the negative log of our concentration of protons H plus. And so we can plug in our negative log of our concentration of H plus which we found to be 0.7747. And this will result in our final answer, which is our ph of the reaction at equilibrium equal to the value 1.11. This would be our final answer to complete this example highlighted in yellow. So I hope that this made sense and let us know if you have any questions.

Consider the unbalanced equation: (d) What pH is required for the reaction to be at equilibrium at 25°C when [I-] = 0.10M and [IO3-] = 0.50 M?

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