Calculate the pH of the solution that results from each
mixture.
...

Question

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

Accepted Answer

Hello. In this problem we are asked to determine the ph of the following solution. We have 25 mL of 250.17 moller this acid with mL of 350.19 molar sodium hypochlorite. Hippo itis acid is a weak acid. So will undergo hydraulic sis will donate a proton to water. What form I bought? I die. Yeah. And her journey minds the hippo I die and I is the conjugate base of our weak acid when we have a weak acid and its conjugate base will form a buffer. And the hippo iodine comes from sodium iodide. Let's consider then what's taking place. We are then combining two solutions for combining 25 mL of 0. Mueller apple itis acid With 35 ml. I have a 0.19 moller sodium iodide solution. So once we combine these two solutions We'll have a total volume. Then of 35 ml plus 25 ml Which is 60 ml. So they're both being diluted. So we can make use of our dilution equation and one V one equals M two V. Two. Both cases were finding the polluted concentration. So M. Two. So the concentration of the sodium hypochlorite. Then after dilution we have our initial concentration of .19 Moeller Times Our Initial Volume 35 ml. We're gonna dilute it then to a final volume of 60 ml. This works out to 0.111 Mueller and then we have our acid Initial concentration of 0.17 moller Initially have 25 the leaders And it gets diluted to 60 ml. So this works out to 0.0708 Molar. So since we have a buffer having a weak acid and its conjugate base, we can make use of the Henderson Hasselbach equation, which tells us that the ph is equal to the PK plus log of concentration of our conjugate base over that of our acid. And the concentration of our nonskid base is going to be equal then to the concentration of our assault. Since the relationship is 1-1 And this was again 0.111 Moeller. So our p will be equal to the negative log of our acid dissociation constant, which is 2.3 times 10 -11. And then we have log of concentration of our why don't you get base right by that of our acid? And this works out to them the ph of 10.83. So the ph of our buffer solution of 10.83 corresponds to answer d. Thanks for watching. Hope this helped

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