I have no idea where to begin with this?? This class is moving so fast! Dr. Bob222 I made a 100 on my prelab!! and the online test was so fast I was going through my notes and I was scrambling to figure it out and look up stuff on the periodic table and all!! But I made an 80! I'm happy with that!
Now... What the heck is this?? PH and acids?
Calculate the concentration of OH− and the pH value of an aqueous solution in which [H3O+] is 0.014M at 25°C. Is this solution acidic, basic or neutral?
Write the acid-dissociation reaction of nitrous acid (HNO2) and its acidity constant expression.
Calculate the concentration of the hydronium ion ([H3O+]) for a solution with the pH = 4.51.
Consider the HF solution with the concentration of 5% (w/v). Calculate its concentration in the molarity. Use four significant figures in your answer.
At equilibrium, it was found that [CH3COOH] = 0.8537 M, [CH3COO−]= 0.0040M, and [H3O+] =0.0040 M. Calculate the acidity constant (Ka) and the pKa value of CH3COOH.
You would do well to post separate posts. Sometimes we don't have time to answer 4-5 questions at a time.
Calculate the concentration of OH− and the pH value of an aqueous solution in which [H3O+] is 0.014M at 25°C. Is this solution acidic, basic or neutral?
pH = -log(H3O^+)
pH = -log(0.014M) etc. It's a matter of finding the right buttons on your computer.
(H3O^+)(OH^-) = Kw = 1E-14. You know Kw and H3O^+, solve for OH^-
If H3O^+ > 10-7M, acid
If H3O^+ = 10^-7M, neutral
If H3O^+ < 10^-7M, basic
To begin with, let's break down each question step by step:
1. Calculate the concentration of OH− and the pH value of an aqueous solution in which [H3O+] is 0.014M at 25°C. Is this solution acidic, basic, or neutral?
To calculate the concentration of OH− and the pH value, you need to use the equation for the ion product of water (Kw):
Kw = [H3O+][OH−]
At 25°C, Kw is approximately equal to 1.0 × 10^−14 mol^2/L^2.
Since the equation is balanced with equal concentrations of H3O+ and OH−, you can assume that [H3O+] = [OH−]. Therefore, you can substitute in 0.014M for [H3O+] and solve for [OH−]. Once you have the concentration of OH−, you can calculate the pH value using the equation pH = -log10[H3O+].
Remember, if the pH value is less than 7, the solution is acidic. If the pH value is greater than 7, it is basic. If the pH value is exactly 7, it is neutral.
2. Write the acid-dissociation reaction of nitrous acid (HNO2) and its acidity constant expression.
The acid-dissociation reaction of nitrous acid (HNO2) can be written as:
HNO2 ⇌ H+ + NO2^-
The acidity constant expression, also known as the acid dissociation constant (Ka), can be written as:
Ka = [H+][NO2^-]/[HNO2]
3. Calculate the concentration of the hydronium ion ([H3O+]) for a solution with the pH = 4.51.
To calculate the concentration of the hydronium ion, you can use the equation for pH:
pH = -log10[H3O+]
Rearranging the equation, you get:
[H3O+] = 10^(-pH)
Plug in the given pH value of 4.51 and calculate the concentration of [H3O+].
4. Consider the HF solution with a concentration of 5% (w/v). Calculate its concentration in molarity. Use four significant figures in your answer.
To calculate the concentration in molarity, you need to convert the given percentage (w/v) into grams of solute per liter of solution.
Assuming the density of the solution is approximately 1 g/mL, you can convert the percentage to grams by multiplying it by the volume of the solution in milliliters. Then divide the mass by the volume in liters to get the concentration in molarity.
Remember to use four significant figures in your answer.
5. At equilibrium, it was found that [CH3COOH] = 0.8537 M, [CH3COO−]= 0.0040M, and [H3O+] =0.0040 M. Calculate the acidity constant (Ka) and the pKa value of CH3COOH.
To calculate the acidity constant (Ka) and the pKa value of CH3COOH, you can use the concentrations of acetic acid ([CH3COOH]), acetate ion ([CH3COO−]), and hydronium ion ([H3O+]) at equilibrium.
The equilibrium equation for the dissociation of acetic acid can be written as:
CH3COOH ⇌ CH3COO^- + H3O+
The acid dissociation constant (Ka) expression can be written as:
Ka = [CH3COO^-][H3O+]/[CH3COOH]
Once you have the acidity constant (Ka), you can calculate the pKa value using the equation pKa = -log10(Ka).
Remember to use the given concentrations at equilibrium to calculate these values.
I hope this helps you get started with the questions about pH and acids! If you have any further questions or need more in-depth explanations, feel free to ask.