The Ka for formic acid is 1.8 x10^-4. What is the pH?
I think what Paul Bogardus is trying to say is, “Not enough information. Without knowing the concentration of the formic acid in the solution, we can’t calculate the pH.” And if it’s not in solution, the question has no meaning.
The Ka for formic acid is 1.8 x10^-4. What is the pH?
I think what Paul Bogardus is trying to say is, “Not enough information. Without knowing the concentration of the formic acid in the solution, we can’t calculate the pH.” And if it’s not in solution, the question has no meaning.
Since both the acids are weak , we can assume that their concentrations after dissociation are almost equal to their initial (undissociated) concentrations. So [H+]^2 =(1.8*10^-4)*0.1+(1.8*10^-5)*0.1 =1.98*10^-5
Since both the acids are weak , we can assume that their concentrations after dissociation are almost equal to their initial (undissociated) concentrations. So [H+]^2 =(1.8*10^-4)*0.1+(1.8*10^-5)*0.1 =1.98*10^-5
No one can tell you. Don’t forget, pH measures the molar concentration of a specific ion, [math]K_a[/math] measures how much an acid dissociates in solution. It’s like asking if the room is green, how many people can sit at the tables?
No one can tell you. Don’t forget, pH measures the molar concentration of a specific ion, [math]K_a[/math] measures how much an acid dissociates in solution. It’s like asking if the room is green, how many people can sit at the tables?
The first thing you need to consider is what is the relationship between pKa and Ka. The second thing you need to consider is that stronger acids have smaller pKa values.
pKa = -log Ka pKa of acetic acid is -log(1.8 × 10^-5) = 4.744
So, as stated in the question, the pKa of the food is 4.09, the pKa of acetic acid is 4.744 so the food is a stronger acid than acetic acid.
The first thing you need to consider is what is the relationship between pKa and Ka. The second thing you need to consider is that stronger acids have smaller pKa values.
pKa = -log Ka pKa of acetic acid is -log(1.8 × 10^-5) = 4.744
So, as stated in the question, the pKa of the food is 4.09, the pKa of acetic acid is 4.744 so the food is a stronger acid than acetic acid.
Yes, pH of 1 is acidic. In fact it’s strongly acidic.
pH = -log10 [H+] therefore a solution of pH =1 would have a hydrogen ion concentration of 0.1 mol/dm3
Yes, pH of 1 is acidic. In fact it’s strongly acidic.
pH = -log10 [H+] therefore a solution of pH =1 would have a hydrogen ion concentration of 0.1 mol/dm3
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The Ka for formic acid is 1.8 x10^-4. What is the pH?
I think what Paul Bogardus is trying to say is, “Not enough information. Without knowing the concentration of the formic acid in the solution, we can’t calculate the pH.” And if it’s not in solution, the question has no meaning.
The Ka for formic acid is 1.8 x10^-4. What is the pH?
I think what Paul Bogardus is trying to say is, “Not enough information. Without knowing the concentration of the formic acid in the solution, we can’t calculate the pH.” And if it’s not in solution, the question has no meaning.
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The dissociation constants of the two acids
K(form)=[ H+]*[HCOO-]/[HCOOH]
K(acet)=[H+]*[CH3COO-]/[CH3COOH]
=> [H+]*{[HCOO-]+[CH3COO-]} = K(form)*[HCOOH]+K(acet)[CH3COOH]
Now, [H+]=[HCOO-]+[CH3COO-]
=> [H+]^2=K(form)*[HCOOH]+K(acet)*[CH3COOH]
Since both the acids are weak , we can assume that their concentrations after dissociation are almost equal to their initial (undissociated) concentrations. So [H+]^2 =(1.8*10^-4)*0.1+(1.8*10^-5)*0.1 =1.98*10^-5
=> [H+]=(1.98*10^-5)^0.5=4.45*10^-3
=> pH= -log4.45*10^-3=2.35
The dissociation constants of the two acids
K(form)=[ H+]*[HCOO-]/[HCOOH]
K(acet)=[H+]*[CH3COO-]/[CH3COOH]
=> [H+]*{[HCOO-]+[CH3COO-]} = K(form)*[HCOOH]+K(acet)[CH3COOH]
Now, [H+]=[HCOO-]+[CH3COO-]
=> [H+]^2=K(form)*[HCOOH]+K(acet)*[CH3COOH]
Since both the acids are weak , we can assume that their concentrations after dissociation are almost equal to their initial (undissociated) concentrations. So [H+]^2 =(1.8*10^-4)*0.1+(1.8*10^-5)*0.1 =1.98*10^-5
=> [H+]=(1.98*10^-5)^0.5=4.45*10^-3
=> pH= -log4.45*10^-3=2.35
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No one can tell you. Don’t forget, pH measures the molar concentration of a specific ion, [math]K_a[/math] measures how much an acid dissociates in solution.
It’s like asking if the room is green, how many people can sit at the tables?
No one can tell you. Don’t forget, pH measures the molar concentration of a specific ion, [math]K_a[/math] measures how much an acid dissociates in solution.
It’s like asking if the room is green, how many people can sit at the tables?
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If you mean Ka value 1.8 x 10^-4 is larger than 4.5 x 10^-5, which would mean formic acid is the stronger acid.
You can check by converting to pKa values. The lower the value the stronger the acid.
pKa = -Iog10Ka
Nitrous acid has pKa of -log10(4.5x10^-5) = -(-4.3) = 4.3
Formic acid has pKa of -log10(1.8x10^-4) = -(-3.7) = 3.7
Formic acid has the lower pKa value
which agrees with formic acid being the stronger acid
Remember Ka and pKa are solvent dependent. e.g. acids have different strengths in water and DMSO as the solvent.
If you mean Ka value 1.8 x 10^-4 is larger than 4.5 x 10^-5, which would mean formic acid is the stronger acid.
You can check by converting to pKa values. The lower the value the stronger the acid.
pKa = -Iog10Ka
Nitrous acid has pKa of -log10(4.5x10^-5) = -(-4.3) = 4.3
Formic acid has pKa of -log10(1.8x10^-4) = -(-3.7) = 3.7
Formic acid has the lower pKa value
which agrees with formic acid being the stronger acid
Remember Ka and pKa are solvent dependent. e.g. acids have different strengths in water and DMSO as the solvent.
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The first thing you need to consider is what is the relationship between pKa and Ka. The second thing you need to consider is that stronger acids have smaller pKa values.
pKa = -log Ka
pKa of acetic acid is -log(1.8 × 10^-5) = 4.744
So, as stated in the question, the pKa of the food is 4.09, the pKa of acetic acid is 4.744 so the food is a stronger acid than acetic acid.
The first thing you need to consider is what is the relationship between pKa and Ka. The second thing you need to consider is that stronger acids have smaller pKa values.
pKa = -log Ka
pKa of acetic acid is -log(1.8 × 10^-5) = 4.744
So, as stated in the question, the pKa of the food is 4.09, the pKa of acetic acid is 4.744 so the food is a stronger acid than acetic acid.
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We DUNNO …. while you have specified [math]K_{a}[/math] for the weak acid … you have not specified the starting molar concentration…
We suppose that you have a [math]0.10•mol•L^{-1}[/math] concentration with respect to formic acid.…
And of course we can represent the dissociation of the acid…
[math]HC(=O)OH(aq) + H_{2}O(l) ightleftharpoons HC(=O)O^{-} + H_{3}O^{+}[/math]
…for which we write the equilibrium expression …. [math]K_{a}=dfrac{[HC(=O)O^{-}][H_{3}O^{+}]}{[HC(=O)OH(aq)]}=1.80×10^{-4}[/math]
And if [math]x•mol•L^{-1}[/math] formic acid dissociates, we can rewrite the expression…
[math]K_{a}=dfrac{x^{2}}{0.1-x}=1.80×10^{-4}[/math]
And if [math]0.1 >> x[/math], then we make the approx…
[math]x_{1}=sqrt{1.80×10^{-4}×0.1}=4.24×10^{-3}[/math]
And we recycle this approximation back into the expression…
[math]x_{2}=sqrt{1.80×10^{-4}×(0.1-4.24×10^{-3})}=4.15×10^{-3}[/math]
[math]x_{3}=sqrt{1.80×10^{-4}×(0.1-4.15×10^{-3})}=4.15×10^{-3}•mol•L^{-1}[/math]
But [math]x=[H_{3}O^{+}][/math], and thus [math]pH=-log_{10}(4.15×10^{-3})=+2.38[/math]...
We DUNNO …. while you have specified [math]K_{a}[/math] for the weak acid … you have not specified the starting molar concentration…
We suppose that you have a [math]0.10•mol•L^{-1}[/math] concentration with respect to formic acid.…
And of course we can represent the dissociation of the acid…
[math]HC(=O)OH(aq) + H_{2}O(l) ightleftharpoons HC(=O)O^{-} + H_{3}O^{+}[/math]
…for which we write the equilibrium expression …. [math]K_{a}=dfrac{[HC(=O)O^{-}][H_{3}O^{+}]}{[HC(=O)OH(aq)]}=1.80×10^{-4}[/math]
And if [math]x•mol•L^{-1}[/math] formic acid dissociates, we can rewrite the expression…
[math]K_{a}=dfrac{x^{2}}{0.1-x}=1.80×10^{-4}[/math]
And if [math]0.1 >> x[/math], then we make the approx…
[math]x_{1}=sqrt{1.80×10^{-4}×0.1}=4.24×10^{-3}[/math]
And we recycle this approximation back into the expression…
[math]x_{2}=sqrt{1.80×10^{-4}×(0.1-4.24×10^{-3})}=4.15×10^{-3}[/math]
[math]x_{3}=sqrt{1.80×10^{-4}×(0.1-4.15×10^{-3})}=4.15×10^{-3}•mol•L^{-1}[/math]
But [math]x=[H_{3}O^{+}][/math], and thus [math]pH=-log_{10}(4.15×10^{-3})=+2.38[/math]...
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