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Ch.18 - Aqueous Ionic Equilibrium

Chapter 18, Problem 41a

Calculate the pH of the solution that results from each mixture. a. 50.0 mL of 0.15 M HCHO2 with 75.0 mL of 0.13 M NaCHO2

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Hey everyone. And welcome back to another video. Calculate the Ph of the solution that results from each mixture. A 50 mL of 0.15 molar formic acid with 75 mL of 0.13 molar sodium format. Where given for answer choices. A 4.70 B 3.86 C 5.67 ND 2.88. For the purposes of this problem, we have to notice that this solution consists of a weak acid H cho two. That's our formic acid as well as its conjugate base. We have sodium format. And the reason why it is the conjugate base is simply because we have to recall that this is an ionic compound and in an aqueous solution, sodium format associates into sodium CS which are ion Spectators and format anions. So now we clearly see that fate and formic acid, they differ by one hydrogen, meaning we have our weak acid and is conjugate base. As a result, we have a buffer solution. And to find the Ph of a buffer solution, we can simply apply the Henderson Hasselback equation which states that Ph is equal to the PK of the weak acid plus the logarithm of base sun of the ratio between the concentrations or the number of moles if we want of the conjugate base. So we can say the number of moles of cho two negative. And we're going to divide that by the number of moles of the weak asset. Now we can use concentrations or moles, it doesn't really matter because we have a ratio right. Concentrations are directly proportional to the number of moles. So what do we need? Well, first of all, we need the PK value of formic acid. And using the constant tables, we notice that the K A value of formic acid is equal to 1.8 multiplied by 10, the negative four. OK. So we have K A and we can easily get our PK value by taking the negative log of K A. Now let's find the number of moles of each species. First of all, let's find the number of moles of the conjugate base. And we know that we can find moles by multiplying volume and molarity. So in this case, our volume is 75 mL notice that it's not really important to convert our unit. So we can skip that step to make it more time efficient. And we are going to multiply that volume by the concentration which is equal to 0.13 molar, we get 9.75 millie mose. Now, the reason why we get mimos is because we didn't convert milliliters to liters. Similarly, we will calculate the number of modes of formic acid. We will use the volume of 50 mL. And we're going to multiply that by the molarity of 0.15 molar, this gives us 7.50 millimeters. And what we notice is that we have everything that we need to solve this problem. We're going to use the equation. We are going to take the negative log of K A in this case, 1.8 multiplied by its sense, the negative force. And we are going to add the log of the ratio that we have. So first of all, we're taking nine point 75 millimoles on top and we're going to divide that by the number of moles of the asset that would be 7.50 millimoles. Now, we can clearly see why we didn't need to perform any unit conversion because our millimals cancel each other out and eventually we arrive at our final answer. So the final answer that we get is 3.86. Essentially what we can say in this case is that the correct answer to this problem is B B 3.86 is the ph of the given solution. Thank you for watching. And I hope to see you in the next one.