Calculate the pH after 0.020 mol of NaOH is added to 1.00 L of each of the four solutions in Exe

Calculate the pH after 0.020 mol of NaOH is added to 1.00 L...

Key Concepts

Acid-Base Reaction Stoichiometry

This concept involves calculating the amounts of acid and base that react based on their mole ratios in the balanced chemical equation. It is essential for determining which reactant is in excess and by how much, particularly when a known amount of strong base is added to an acid solution.

pH Calculation

Calculating pH requires determining the concentration of hydrogen ions in the solution after the reaction has reached completion or equilibrium. For a basic solution, this might involve calculating the hydroxide ion concentration and then converting to pH, while for an acidic or buffered system, it involves the residual acid or the ratio of weak acid to conjugate base.

Buffer Systems

When a weak acid is partially neutralized by a strong base, the resulting mixture contains significant amounts of both the weak acid and its conjugate base, forming a buffer solution. The pH of a buffer is usually calculated using the Henderson-Hasselbalch equation, which relates pH to the pKa and the ratio of the concentrations of the conjugate base and weak acid.

Titration of Acids and Bases

Titration is a technique used to determine the concentration of an acid or base by gradually adding a titrant of known concentration until the reaction reaches an equivalence point. Understanding titration curves, especially at points before or after the equivalence point, is crucial for predicting how the pH will change upon addition of a strong base.

Transcript

00:01 This problem asks us to look at results from a previous problem and consider what the ph would be if we added 0 .020 moles per liter of solution of sodium hydroxide.

00:14 And the first thing we're adding, our first solution was 0 .10 molar, propanoic acid, and we're adding 0 .020 molar, n .a .h.

00:27 Okay, so let's write a problem and get it solving.

00:36 We are given hc3h502, and we're adding hydroxide to that, and we will produce the anion and water.

00:57 And water, of course, will ignore.

01:01 So our initial concentration is 0 .10 -molar and 0 .020 molar and 0 .0 .000 molar.

01:16 Will react.

01:18 It's clearly limiting the point.

01:24 So this will go down by that amount and this will increase by 0 .020, giving us an equilibrium of 0 .080, 0 and 0 .020.

01:50 Now we can go ahead look at this and then we're going to look at a somewhat different equation.

01:59 I'm going to colors here and this time we're going to look at the following this is what we're going to be looking at next and we're going to do another ice this is 0 .080 this is 0 .0 .0 .020 my original h plus is 0 is 0 .0 my original h plus is zero so this will be increased by x increased by x decrease by x and then we can write our k a expression i guess for this one no we can do k a and i've already done for these for all these so let me just go ahead and write 1 .3 times 10 to the minus 5 that's given as our k a way back in our old problem equals 0 .020 plus x times x over 0 .0 .0 .0 we're going to ignore the plus x and minus x right there because they're pretty insignificant.

03:55 So this will reduce if you will to 0 .020 x over 0 .080.

04:09 If we solve our math on this one, x is equal to 5 .2 times 10 to the minus 5.

04:26 And that's the molarity of our h plus.

04:32 To find the ph, we're going to take the negative log of 5 .2 times 10 to the minus 5.

04:44 And we get 4 .28 is the ph of solution 1.

04:55 That's the answer for the ph of solution 1.

05:00 Solution 2 is 0 .10 molar sodium propanoid.

05:08 And on solution 2, this is going to be pretty simple here, we've got 0 .0, whoops, since we have 0 .10 molar c3h502, and that's minus, and that's our conjugate base, we have 0 .020 molar oh, we're going to figure out what our oh comes.

05:58 Concentration is here.

06:06 And this is going to be reduced by 0 .10 -0 molar.

06:22 I'm trying to think you hear what i'm going to have.

06:30 I feel like i need to think about this one a little bit more.

06:38 If i'm adding this malarity, yep, that's going to be my, i'm going to take my p -o -h isn't equal to the negative log of 0 .020 molar equals 1 .7.

06:58 So our ph is 14 .00 minus 1 .70 equals 12 .30 molar.

07:13 And for number three, that's our super simple one.

07:25 We have pure water...

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