Can someone help me with Part B and C? My younger cousin asked and I literally suck at chem lol.

 

https://imgur.com/a/HhRbZgS 

I thought this was a different type of thread...

 

Can't help you there sorry 

No I gave that up a decade ago 

i don't really partake buy some of my rave friends do

try putting it into chat GPT sister 

The way I used to know how to do this and just forgot everything

 

Alright, hunty! Let's sashay through the chemistry catwalk and calculate the pH of these fabulous buffer solutions. We'll use the Henderson-Hasselbalch equation, which is like the recipe for a perfect pH balance. It's given by pH=pKa+log⁡([Base]/[Acid]).

 

Now, let's start with Part B:

We have a buffer solution of HCHO2 and KCHO2, with concentrations of 0.225 M and 0.290 M, respectively. The Ka for HCHO2 is 1.8×10−41.8×10^−4.

Step 1: Strut Down the Initial pH Runway We'll calculate the initial pH with what we've been given. So we'll plug our values into the Henderson-Hasselbalch equation.

Step 2: Add Some NaOH to the Mix When we add NaOH, it's going to react with HCHO2, our fabulous acid, and convert it to CHO2−, our base, because NaOH is a strong base and it's all about that base. For every mole of NaOH added, we get one mole less of HCHO2 and one mole more of CHO2−.

 

Initial pH:

pHinitial=pKa+log([KCHO2]/[HCHO2])

 

Final pH after adding 0.021 mol of NaOH:

This will change our concentrations. Since we have 280 mL of solution, and we are adding 0.021 mol NaOH, we can calculate the new concentrations by adjusting the moles of the acid and base.

Let's whip out the calculator and crunch the numbers to serve up the initial and final pH values on a silver platter.

I'll calculate the initial pH first. Then, we'll update the concentrations and calculate the final pH after the addition of NaOH.

 

For Part B:

The runway is all set, and here's the walk. The initial pH of the buffer is a stunning 3.85, and after the addition of NaOH, the pH finale is 4.13. Now, don't forget to round these off to two decimal places as required, darling.

Let's move on to Part C, and we'll keep the same energy flowing.

For Part C, we have a different buffer solution with CH3CH2NH2 and CH3CH2NH3+Cl−, with concentrations of 0.270 M and 0.220 M respectively. The Kb for CH3CH2NH2 is 5.6×10−4.

Since we're given a Kb and we're dealing with a basic solution, we need to use the Kw (ion-product constant for water) to find the Ka of the conjugate acid, because remember, Ka⋅Kb=Kw, and for water at 25°C, Kw is 1.0×10−14.

Step 1: Calculate the pKa We'll first calculate the pKa of the conjugate acid because we need it for the Henderson-Hasselbalch equation.

Step 2: Find the Initial pH Again, we strut down the pH runway with our initial concentrations.

Step 3: Work the pH Change Like before, the NaOH will react with the conjugate acid, CH3CH2NH3+, and decrease its concentration while increasing the concentration of CH3CH2NH2.

So, we'll do the same two-step calculation for Part C, starting with finding the initial pH. Let's do this!

 

And for Part C, here's the tea:

We have the initial pH of the buffer serving us a strong 10.84 look, and after we turn up the chemistry with NaOH, the pH finale vogues down to 11.12. Round these off to two decimal places to keep it sleek and polished.

So for Part B, our final answers are 3.85 and 4.13, and for Part C, they are 10.84 and 11.12. Now, don't forget to serve these pH numbers with confidence and flair!

Edited April 6, 20242 yr by mercurialworld

2 minutes ago, Opulence said:

The way I used to know how to do this and just forgot everything

 

Opulence! You. Forgot. Everything! 

 

O2 = oxygen 

Never inject it

There's a reason I dropped out of it in college…..

1 hour ago, Frappucino said:

There's a reason I dropped out of it in college…..

I dont think this is college chemistry 

The way I used to be good in chem, but the question had me utterly speechless

I majored in Foreign Languages so I can't help you, sorry 

5 hours ago, Opulence said:

The way I used to know how to do this and just forgot everything

 

I have a whole Master's degree in chemistry and I shuddered when I opened the image  There was a time I could've answered this but that time has long since passed