Find the Ph of a 0.0050 M Sulfuric Acid Solution.
what is the PH of a 0.00050M solution of H2SO4?
Best Answer
+109 H2SO4 is classified as a "strong" acid, which means it dissociates completely. Knowing this, we can surmise that the molarity of hydrogen ions will be the same as the molarity of our given acidic solution (H2SO4), at 0.0005M.
$$[H_2SO_4] = 0.0005M$$
Dissociates to equal concentrations of the ions H+, and HSO4 -:
$$[H^+] = 0.0005M$$
$$[HSO_4^-] = 0.0005M$$
Thus the molarity of hydrogen atoms up to this point is 0.0005M, giving a pH = -log([H+]) = -log(0.0005) = 3.
Now, we need to calculate the molarity of hydrogen dissociated from the decidedly "weak" acidic ion HSO4 -. Since HSO4 - is a weak acid, we need to find and use the dissociation constant of HSO4 - to calculate the molarity of H+ from the parent compound.
$$Ka = [H^+][SO_4^2^-]/[HSO_4^-] = 1.1 x 10^-^2$$
Solving for Ka will give us our desired hydrogen molarity, so consider the following:
$$([H^+]*[SO_4^2^-]) / [HSO_4^-] = 1.1 x 10^-^2$$
We know that [H+] = [SO4 2-] = 0.0005M so we can consider those, respectively, as a variable, "x" for example. We also know that the concentration of [HSO4 -] will be it's curent concentration MINUS the concentration of those that dissociated further to the ions [H+] and [SO4 2-]; giving us:
$${\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{0.000\: \!5}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}} = {\mathtt{1.1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}$$
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.011}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{0.000\: \!005\: \!5}} = {\mathtt{0}}$$
Solving for "x", we'll get:
$$x = [H^+] = 0.00047913$$
Which indicates our [H+] concentration for the second hand dissociation of H2SO4, or the first hand dissociation of HSO4 -. To get the total concentration of [H+] ions in the solution, you simply add them together and run the resulting number through the pH formula.
$$0.0005M + 0.00047913M = 0.00097913M$$
$$pH = -log([H^+]) = -log(0.00097913) = 3.009 \approx 3.01$$
Hope this helps. Thank you Melody, and CPhil, for pointing out my mistake. I've corrected it now. :)
+109 H2SO4 is classified as a "strong" acid, which means it dissociates completely. Knowing this, we can surmise that the molarity of hydrogen ions will be the same as the molarity of our given acidic solution (H2SO4), at 0.0005M.
$$[H_2SO_4] = 0.0005M$$
Dissociates to equal concentrations of the ions H+, and HSO4 -:
$$[H^+] = 0.0005M$$
$$[HSO_4^-] = 0.0005M$$
Thus the molarity of hydrogen atoms up to this point is 0.0005M, giving a pH = -log([H+]) = -log(0.0005) = 3.
Now, we need to calculate the molarity of hydrogen dissociated from the decidedly "weak" acidic ion HSO4 -. Since HSO4 - is a weak acid, we need to find and use the dissociation constant of HSO4 - to calculate the molarity of H+ from the parent compound.
$$Ka = [H^+][SO_4^2^-]/[HSO_4^-] = 1.1 x 10^-^2$$
Solving for Ka will give us our desired hydrogen molarity, so consider the following:
$$([H^+]*[SO_4^2^-]) / [HSO_4^-] = 1.1 x 10^-^2$$
We know that [H+] = [SO4 2-] = 0.0005M so we can consider those, respectively, as a variable, "x" for example. We also know that the concentration of [HSO4 -] will be it's curent concentration MINUS the concentration of those that dissociated further to the ions [H+] and [SO4 2-]; giving us:
$${\frac{\left({{\mathtt{x}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{0.000\: \!5}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}} = {\mathtt{1.1}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{2}}}$$
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.011}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{0.000\: \!005\: \!5}} = {\mathtt{0}}$$
Solving for "x", we'll get:
$$x = [H^+] = 0.00047913$$
Which indicates our [H+] concentration for the second hand dissociation of H2SO4, or the first hand dissociation of HSO4 -. To get the total concentration of [H+] ions in the solution, you simply add them together and run the resulting number through the pH formula.
$$0.0005M + 0.00047913M = 0.00097913M$$
$$pH = -log([H^+]) = -log(0.00097913) = 3.009 \approx 3.01$$
Hope this helps. Thank you Melody, and CPhil, for pointing out my mistake. I've corrected it now. :)
+115420 I do not know if it is correct but I will give you 3 points for your efforts
+121064 You lost me at the word, "dissociates"......The rest???.....I'll take your word for it......
(I'll give you 3 points even if you made up a good story.....at least it looks impressive)
+109 I've corrected it now, thank you. :) I took the wrong approach, and didn't fully complete the question. I gave him bounds by accident instead of following through.
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Find the Ph of a 0.0050 M Sulfuric Acid Solution.
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