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I'm trying to solve a simple problem which is driving me crazy. It says:

Estimate the pH of a solution that contains 1 gram of $\ce{H_2SO_4}$ dissolved in 1 liter of water.

When I solve the problem, I dissociate the $\ce{H_2SO_4}$ like this: $$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}$$

And I get a pH of around 1.99

However, following my book's solution, the dissociation occurs as follows: $$\ce{H2SO4 <=> 2H+ + {SO4}^2-}$$ And they say the pH is around 1.69

Now, my question is: why is my dissociation equation wrong?

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  • $\begingroup$ There's nothing wrong with it. $\endgroup$ – Avnish Kabaj Jun 22 '18 at 18:53
  • $\begingroup$ Nothing wrong with my solution (pH=1.99), or with my book's solution (pH=1.69)? $\endgroup$ – Jose Lopez Garcia Jun 22 '18 at 18:54
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    $\begingroup$ It's diluted acid so both protons are dissociated. Also this thing was done to death... $\endgroup$ – Mithoron Jun 22 '18 at 18:55
  • $\begingroup$ I don't understand the downvote, I'm beginning with Chemistry, I find this subject very hard, and on top of that I'm afraid of asking here because of downvotes. I don't know who else to ask to be honest. $\endgroup$ – Jose Lopez Garcia Jun 22 '18 at 18:57
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    $\begingroup$ Don't worry about the downvotes too much every new users gets a couple before they learn the ropes. Most likely it got downvoted because it's been marked as a duplicate. As for my earlier comment I meant that your dissasociation equation is correct but there will be another equation too $$\ce{HSO4- <=> H+ + SO4^{2^-}}$$ $\endgroup$ – Avnish Kabaj Jun 22 '18 at 19:05
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Your problem is that you only accounted for the first dissociation of $\ce{H2SO4}$, a polyprotic acid -- your book needed the extra specificity from the second dissociation. I will walk through the entire process, including the parts that you know already.

Begin by finding the molar mass of $\ce{H2SO4}$ in order to find out to how many moles one gram of it is equivalent. Then, convert to molarity (concentration) using the given volume of water.

$$\ce{MM_{H_2SO_4} = 2*1.01 g + 1*32.06 g + 4*16.00 g = 98.08 g}$$

$$\ce{\frac{1 g H2SO4}{1}\times\frac{1 mol H2SO4}{98.08 g H2SO4} = 1.0\times10^{-2} mol H2SO4}$$

$$\ce{\frac{1.0\times10^{-2} mol H2SO4}{1 L H2O} = 1.0\times10^{-2} M H2SO4}$$

Although the ICE-box is a formality for such a strong acid, it can still be shown.

\begin{array}{|c|c|c|c|c|} \hline \text{Initial}:& 1.0\times10^{-2} & & 0 & 0 \\ \hline & \ce{H2SO4} & \ce{H2O} & \ce{H3O+} & \ce{HSO4-}\\ \hline \text{Change}: & -x & & +x & +x \\ \hline \text{Equilibrium}: & 0 & & 1.0\times10^{-2} & 1.0\times10^{-2}\\ \hline \end{array}

The second ICE-box is a good way of organizing the second dissociation. Transfer the equilibrium concentrations from the first table. All calculations up to the line are for finding the change (using $\ce{K_{a(2)} = 1.2\times10^{-2}}$). Note that after $y$ is found, it is used again in the second ICE-box to determine the equilibrium concentrations after the second dissociation. Also note that you cannot neglect the $y$ after the second equation because of the similar magnitudes of the molarity and the $K_a$ and must use the quadratic formula.

\begin{array}{|c|c|c|c|c|} \hline \text{Initial}:& 1.0\times10^{-2} & & 1.0\times10^{-2} & 0 \\ \hline & \ce{HSO4-} & \ce{H2O} & \ce{H3O+} & \ce{SO4^{2-}}\\ \hline \text{Change}: & -y & & +y & +y \\ \hline \text{Equilibrium}: & 0.5\times10^{-2} & & 1.5\times10^{-2} & 4.8\times10^{-3}\\ \hline \end{array}

$$\ce{K_a = \frac{[H3O+][SO4^{2-}]}{[HSO4-]}}$$

$$\ce{1.2\times10^{-2} = \frac{(1.0\times10^{-2} + y)(y)}{1.0\times10^{-2} - y}}$$

$$\ce{1.2\times10^{-4} - (1.0\times10^{-2})y = (1.0\times10^{-2})y + y^2}$$

$$\ce{1.2\times10^{-4} = (2.0\times10^{-2})y + y^2}$$

$$\ce{0 = y^2 + (2.0\times10^{-2})y - 1.2\times10^{-4}}$$

\begin{split} \ce{y} & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ & = \frac{-(2.0\times10^{-2})\pm\sqrt{(2.0\times10^{-2})^2-4(1)(-1.2\times10^{-4})}}{2(1)}\\ & = \frac{-2.0\times10^{-2}\pm\sqrt{4.0\times10^{-4}+4.8\times10^{-4}}}{2}\\ & = \frac{-2.0\times10^{-2}\pm\sqrt{8.8\times10^{-4}}}{2}\\ & \approx 4.8\times10^{-3} \end{split}


Plug into the p function to determine the pH.

$$-\log(1.5\times10^{-2})=1.82$$

Note that $-\log(2\times10^{-2})=1.69$ so your book probably rounded to one significant figure (which would make sense given the way the problem is worded).

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