Succinic acid (H2 C4 H6 O4), which we will denote H2 Suc is a biologically relevant diprotic aci

Succinic acid (H2 C4 H6 O4), which we will denote H2 Suc is...

Key Concepts

Salt Hydrolysis

Salt hydrolysis occurs when the salt derived from a weak acid or weak base interacts with water, leading to the formation of acidic or basic solutions. Understanding how the conjugate base of a weak acid, when paired with a neutral cation, reacts with water to form OH? or H? ions allows for predicting whether the resulting solution will be acidic, neutral, or basic.

Polyprotic Acids

Polyprotic acids are acids that can donate more than one proton (H?) per molecule. They dissociate in a stepwise manner with each dissociation characterized by its own acid dissociation constant (Ka). Understanding the sequential deprotonation process helps in evaluating the overall acid behavior and determining how each step contributes to the final pH of a solution.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, quantifies the strength of an acid in solution by representing the equilibrium between the undissociated acid and its dissociated ions. A small Ka value indicates a weak acid that only partially dissociates in water, which is key to calculating equilibrium concentrations and estimating pH in acidic solutions.

pH Calculation for Weak Acids

Calculating the pH of a weak acid solution involves setting up an equilibrium expression based on the acid's dissociation constant. Approximations are often applied, especially when the degree of dissociation is small relative to the initial concentration, simplifying the calculation process. This methodology is essential for analyzing solutions in which the acid does not fully ionize.

Transcript

00:01 Chapter 16, promen 108, they're saying that we have a diaproduct acid, which is going to be represented as h2 succinate, and it's a biological relevant diapuric acid.

00:13 And so they gave us our k -a -1 and k -a -2 values, and they asked us to do some things with the information given.

00:21 And so for part a, it acts to determine the ph of a 0 .3 -2 molar solution of h -2 -sux at 25 degrees celsius.

00:31 And only the first dissociation is relevant.

00:34 So what we first had to do is right out of our equation of what's going to happen during this dissociation.

00:41 So we have h2 sucks, and it dissociates into our h plus and h sucks minus.

00:50 Remember that all these solutions are equates and will be included in our ice table.

00:56 So now that we do our ice table, we know that we have a starting concentration of 0 .32, and then zero for our products.

01:04 And then we're going to then subtract from the initial concentration and then add to the other side.

01:10 For the equilibrium, you just add up your first two rows for the initial and the change column.

01:16 And so we're going to get 0 .32 minus x, and then x.

01:23 So we're already given our ka value, but we need to remember that ka is equal to our products over reactives.

01:30 So our general equation for ka should be our h plus, ion concentration times our h sucks minus concentration all over the h2 sucks concentration and so since we see that we can then plug in our values that we have and since h plus and h sucks minus i'll have an x it's just going to be x squared and then we're going to have 0 .3 to minus x and our k a for ka1 since we only dissociated it once is equal to 6 .9 times 10 to negative 5th.

02:11 And since the 6 .9 times 10 to negative 5th is such a small number compared to our initial concentration, the x in the bottom is measurable, and we do not have to include it, which also makes our calculations a little bit easier.

02:24 So to find our x, we have x squared is equal to the 6 .9 times 10 to the negative 5th times 0 .32.

02:36 And then you just take the square root of.

02:38 Both sides and our x ends up being 4 .7 times 10 to negative 3 and so it asks us what is the concentration of um it's asking us to determine the ph of the concentration of 0 .32 so this is our x x is equal to 4 .7 times 10 to negative 3 and we know that x is equal to our h plus so to find our ph, we're just going to do the negative log of our x concentration.

03:13 And when we plug that in, we get the value of 2 .33.

03:32 And that's what our ph is equal to.

03:34 So that's for the answer for part a.

03:37 Now for part b, it is asking us to determine the molar concentration of just sucks two minus.

03:43 So we're going to have to go through another dissociation of the first equation that we just did with our h sucks.

03:50 And then we also going to have an initial concentration of h plus, which is going to be the 4 .7 times 10 to the negative third.

03:59 So when we read out this equation, we're going to start off with our h -sup minus, and that's going to dissociate into our h -plus plus succinate to minus.

04:11 All right.

04:12 So now since we have this equation, we are going to write our ice table.

04:15 And like how i said earlier, we're going to just use the same, that we had in the final row of that for both our h sucks minus plus h plus.

04:24 The reason why we're doing this is because it's a continuation of the same problem.

04:29 It's just asking for the next dissociation constant.

04:32 So we already created some h plus ions in the first part, and we also created the h sucks minus.

04:39 So the only thing that was not produced was the sucks 2 minus.

04:42 So now we're then just going to subtract x from the reactant side and then add it to the products.

04:48 And then for the equilibrium, we just add those two rows up.

04:52 So we get 4 .7 times 10 to negative 3 minus x, and then 4 .7 times 10 to negative 3 plus x and then x.

05:04 So what we said before, our ka is equal to our products over reactants.

05:10 And when we use what our equation is giving us, we should have our h plus concentration times our succulent 2 minus concentration.

05:21 Over our h sucks minus concentration.

05:26 And when we plug in our values that we are given in our equilibrium part, we are getting 4 .7 times 10 to negative 3 plus x times x over 4 .7 times negative 3 minus x.

05:45 We also know our k2 value is going to be 2 .5 times 10 to negative 6.

05:52 One thing to note here, once again, is that since our k -a -2 value is so small in comparison to our initial concentration, those xs are negligible and we don't need to include them.

06:06 So our new equation is going to be 2 .5 times 10 to negative 6 is equal to 4 .7 times 10 to negative 3x over 4 .7 times 10 to negative 3x...

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