#1**+10 **

H_{2}SO_{4} 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 (H_{2}SO_{4}), at 0.0005M.

$$[H_2SO_4] = 0.0005M$$

Dissociates to equal concentrations of the ions H^{+}, and HSO_{4}^{-}:

$$[H^+] = 0.0005M$$

$$[HSO_4^-] = 0.0005M$$

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 HSO_{4}^{-}. Since HSO_{4}^{-} is a weak acid, we need to find and use the dissociation constant of HSO_{4}^{-} 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^{+}] = [SO_{4}^{2-}] = 0.0005M so we can consider those, respectively, as a variable, "x" for example. We also know that the concentration of [HSO_{4}^{-}] will be it's curent concentration MINUS the concentration of those that dissociated further to the ions [H^{+}] and [SO_{4}^{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 H_{2}SO_{4}, or the first hand dissociation of HSO_{4}^{-}. 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. :)

Sorasyn Dec 5, 2014

#1**+10 **

Best Answer

H_{2}SO_{4} 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 (H_{2}SO_{4}), at 0.0005M.

$$[H_2SO_4] = 0.0005M$$

Dissociates to equal concentrations of the ions H^{+}, and HSO_{4}^{-}:

$$[H^+] = 0.0005M$$

$$[HSO_4^-] = 0.0005M$$

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 HSO_{4}^{-}. Since HSO_{4}^{-} is a weak acid, we need to find and use the dissociation constant of HSO_{4}^{-} 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^{+}] = [SO_{4}^{2-}] = 0.0005M so we can consider those, respectively, as a variable, "x" for example. We also know that the concentration of [HSO_{4}^{-}] will be it's curent concentration MINUS the concentration of those that dissociated further to the ions [H^{+}] and [SO_{4}^{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 H_{2}SO_{4}, or the first hand dissociation of HSO_{4}^{-}. 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. :)

Sorasyn Dec 5, 2014

#2