Calculate the pH and [S^2-] in a 0.10-H2 S solution. Assume Ka1=1.0 ×10^-7 ; Ka2=1.0 ×10^-19.

Name: Problem 109, Chapter 13: Chemistry An Atoms First Approach
Uploaded: 2020-05-07T16:16:26-08:00
Duration: 3 min 13 s
Channel: Anatole Borisov
Description: Calculate the pH and [S^2-] in a 0.10-H2 S solution. Assume Ka1=1.0 ×10^-7 ; Ka2=1.0 ×10^-19.

Calculate the pH and [S^2-] in a 0.10-H2 S solution. Assume...

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium is the study of how acids and bases interact in solution, ultimately establishing an equilibrium between different species such as undissociated acids, conjugate bases, and hydrogen ions. This concept is essential for understanding how the concentration of hydrogen ions determines the pH of a solution, and it forms the basis for calculating pH in systems involving weak acids and bases.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, quantifies the extent to which a weak acid donates protons to water, forming hydrogen ions and its conjugate base. A small Ka value indicates a weak acid that only partially dissociates in solution, which is critical when setting up equilibrium expressions and approximations for pH calculations.

Stepwise Dissociation in Polyprotic Acids

Polyprotic acids, such as diprotic acids, lose protons in a sequential, stepwise manner, each with its own dissociation constant. Understanding this concept is crucial for correctly setting up and solving multiple equilibrium expressions for each proton loss, as the equilibrium of one step can influence the concentrations in subsequent steps.

pH Calculation from Equilibrium Concentrations

Calculating pH from a weak acid involves using established equilibrium expressions to determine the concentration of hydrogen ions in solution. The pH is then found using the relationship pH = -log[H+]. This process typically involves solving equilibrium equations and may require approximations when the acid is very weak or when sequential deprotonation steps occur.

Equilibrium Approximations

When dealing with weak acids with small dissociation constants, the change in concentration due to dissociation is often minimal compared to the initial concentration. This allows for simplifications in the equilibrium calculations, such as assuming the initial concentration remains largely unchanged, which simplifies the mathematical process of determining species concentrations.

Transcript

00:01 So this question here is asking about the two dissociations of h2s, which the first of these is h2s plus h2o, becoming h3o plus s plus sh minus.

00:13 And the second of these is sh minus plus water being in equilibrium with h3o plus and s2 minus.

00:28 And so what we're asked to do here, we're asked to figure out what the concentration of s2 minus is, and we're also asked to find the ph, which is another way of saying, what's the concentration of h3o plus? and so to do this first, we have to first just approach this first equilibrium.

00:46 And so we're going to treat this like we always have, where with ka equal to the concentration of h3o plus times the concentration of sh minus, over the initial concentration of h2s, which is equal to x squared over the initial concentration, which we are told is 0 .1 minus x.

01:12 We ignore x because it's much smaller than 0 .1.

01:14 This is from an ice table.

01:16 If you don't understand what this is coming from, write out an ice table for this dissociation and convince yourself that this is what ends up.

01:23 And so we're going to using this 10 to negative seventh value for ka, we'll find that x equals 10th to negative fourth.

01:35 And so now what we do is we can actually figure out what the ph is.

01:41 So because going back to this first slide here, ka2 is a lot, lot smaller than ka1, 10 to negative 19th versus 10 to negative 7th.

01:49 There will be a tiny amount of extra h3o plus added whenever the second association happens, but it's basically negligible.

01:58 And so what we really can do is we can say that since this is equal to the concentration of h3o plus, if we take the negative log of that, so ph equals the negative log of x, we will get our value for the ph, which is 4...

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