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What is the pH of $10^{-8}~\mathrm{M}$ $\ce{HCl}$ solution in water?

My attempt:

pH = $-\log(10)^{-8}$ = 8

But this is wrong because it should be acidic. Where have I gone wrong?

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It is known that the equilibrium constant for the following reaction is $10^{-14}$.

$$\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}$$

That means,

$$\frac {[\ce{H3O+(aq)}][\ce{OH-(aq)}]} {[\ce{H2O(l)}][\ce{H2O(l)}]} = 10^{-14}$$

where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$.

Initially, we have $[\ce{H3O+(aq)}] = 10^{-8}$.

Then, assuming that $x$ moles of $\ce{H3O+(aq)}$ is generated. An equal amount of $\ce{OH-(aq)}$ must also be generated:

$$(10^{-8}+x)(x) = 10^{-14}$$

Solving for $x$ gives $x = 9.51 \times 10^{-8}$.

So, the total concentration of $\ce{H3O+(aq)}$ is $1.051\times10^{-7}$.

Therefore, the pH is $6.978$.


The problem with your approach

Let the initial molarity of the hydronium ion be $a$.

Solving $(a+x)(x) = 10^{-14}$, if $a$ is well above $10^{-5}$, gives $x \approx 0$.

The approximation that the pH is equal to $-\log{{\left([\ce{HCl(aq)}]\right)}}$ is only accurate for normal concentrations.

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    $\begingroup$ I'll quibble. ...where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$. just isn't true. The assumption is that the concentration of water is a constant and is thus ignored in the autopyrolysis equation for water. Water is about 55.5 molar of course. $\endgroup$ – MaxW Jan 22 '17 at 23:31
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    $\begingroup$ Here is another approach: Suppose we have a 10^-8 M HCl solution. We have: Mass balance: [H3O+] = [Cl-] = 10^-8 (M). Charge balance: [H3O+] = [Cl-] + [OH-]. Autoprotolysis of water: [H3O+] * [OH-] = 10^-14. We multiply the charge balance with [H3O+] and get: [H3O+]^2 = [H3O+] * [Cl-] * [H3O+] * [OH-] = [Cl-] * [H3O+] + 10^-14 => We insert the value of [Cl-] and get: [H3O+]^2 = 10^-8 * [H3O+] + 10^-14 => [H3O+]^2 - 10^-8 * [H3O+] - 10^-14 = 0 We solve the equation for [H3O+] and get: [H3O+] = 1,0512*10^-7 => pH = 6.98 ≈ 7 $\endgroup$ – Bive Mar 21 '17 at 21:59
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Caution!

This answer uses a simplification! It assumes that the concentration of $\ce{H3O+}$ as contributed by $K_\mathrm{w}$ is constant at $10^{-7}~\mathrm{M}$. This is incorrect. In reality, an additional equation must be introduced and the resulting formula is cubic rather than quadratic. Unfortunately, solving a cubic equation is very non-trivial and I am not able to do it. One possible cubic equation with which the final value should be calculateable is the following:

$$x^3 + K_\mathrm{a}x^2 - (K_\mathrm{a}c_0)x - K_\mathrm{a}K_\mathrm{w} = 0\\ x = [\ce{H3O+}]$$

Due to the approximation, this answer is off the actual value. The other answer uses a different simplification which arrives at a result closer to the actual value. I’m leaving this post here for posteriority, because it still serves to exemplify how complicated the calculation must be at least and how easily a pitfall is encountered.

Original answer

You used the simplified equation $(1)$ wherein acid is a strong, monoprotic acid. It encompasses a few shortcuts that are only valid for reasonably concentrated strong acids. Using it for an acid this dilute will break a lot of the assumptions used. To clarify, I will use the long method.

$$\mathrm{pH} = -\lg c_0(\text{acid})\tag{1}$$

Upon dissolution of any acid (using hydrogen chloride as an example, but any will work) equilibrium $(2)$ will be found.

$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag{2}$$

This equilibrium can be adequately described by the acidity constant $K_\mathrm{a}$ which is defined as shown in equation $(3)$ (note that where concentrations are implied, activities are actually used. The solvent does not show up because its activity is $1$).

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{HCl}]}\tag{3}$$

The most important simplification present in $(1)$ is that it assumes $c_0(\ce{H3O+}) \ll c_0(\ce{HCl})$ or — equally valid — $c_\mathrm{initial}(\ce{H3O+}) \ll c_\mathrm{final}(\ce{H3O+})$. In that case, it can be equated that $[\ce{H3O+}] \approx [\ce{Cl-}]$ which is used to reduce the number of independent variables and solve the equation. This will not be the case in this calculation with $c_0(\ce{HCl}) = 10^{-8}~\mathrm{M}$ and therefore we need to explore other methods. What we can say is shown in $(4)$.

$$\begin{align}[\ce{Cl-}] + [\ce{HCl}] &= c_0(\ce{HCl})\\ c_0(\ce{H3O+}) + [\ce{Cl-}] &= [\ce{H3O+}]\end{align}\tag{4}$$

$c_0(\ce{H3O+})$ is known from the autodeprotonation of water $(5)$ from which follows that $c_0(\ce{H3O+}) = 10^{-7}~\mathrm{M}$.

$$\begin{align}\ce{2 H2O <=> H3O+ + OH-}&&K_\mathrm{w} = 10^{-14}\end{align}\tag{5}$$

Now, finally, we have enough information to solve the equation.

$$\begin{align}K_\mathrm{a} &= \frac{(10^{-7} + [\ce{Cl-}])[\ce{Cl-}]}{c_0(\ce{HCl}) - [\ce{Cl-}]}\tag{6.1}\\[0.8em] K_\mathrm{a} &= \frac{10^{-7}\times [\ce{Cl-}] + [\ce{Cl-}]^2}{c_0(\ce{HCl}) - [\ce{Cl-}]}\tag{6.2}\\[0.8em] K_\mathrm{a} c_0(\ce{HCl}) - K_\mathrm{a} [\ce{Cl-}] &= 10^{-7}[\ce{Cl-}] + [\ce{Cl-}]^2\tag{6.3}\\[0.6em] 0 &= [\ce{Cl-}]^2 + (10^{-7} + K_\mathrm{a})[\ce{Cl-}] - K_\mathrm{a} c_0(\ce{HCl})\tag{6.4}\\[0.6em] [\ce{Cl-}]_{1/2} &= \frac{-K_\mathrm{a} - 10^{-7} \pm \sqrt{(10^{-7} + K_\mathrm{a})^2 + 4K_\mathrm{a} c_0(\ce{HCl})}}{2}\tag{6.5}\end{align}$$

Plugging in all the values (taking $K_\mathrm{a}$ from Wikipedia) and realising that only the positive square root makes sense, this is our result:

$$\begin{align}[\ce{Cl-}]_{1} &= \frac{-10^{5.9} - 10^{-7} + \sqrt{(10^{-7} + 10^{5.9})^2 + 4 \times 10^{5.9} \times 10^{-8}}}{2}\tag{6.6}\\[0.8em] [\ce{Cl-}]_1 &= \frac{-10^{5.9} - 10^{-7} + \sqrt{(10^{-7} + 10^{5.9})^2 + 4 \times 10^{-2.1}}}{2}\tag{6.7}\\[0.6em] [\ce{Cl-}]_1 &= 10^{-8}\tag{6.8}\end{align}$$

Thankfully, this equates to full deprotonation as the theory predicts. In the simplified context $(1)$, we just assume that complete deprotonation occurs. We can plug the value of $[\ce{Cl-}]$ into equation $(4)$ to get the final result:

$$\begin{align}c_0(\ce{H3O+}) + [\ce{Cl-}] &= [\ce{H3O+}]\tag{4}\\[0.4em] 10^{-7}~\mathrm{M} + 10^{-8}~\mathrm{M} &= 1.1 \times 10^{-7}~\mathrm{M}\tag{7}\\[0.4em] \mathrm{pH} = -\lg [\ce{H3O+}] &= 7\lg 1.1 = 6.959\tag{8}\end{align}$$

You experience a very slight decrease in $\mathrm{pH}$ but the principal contributor of $\ce{H3O+}$ ions is still the autoprotonation of water.

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    $\begingroup$ For pH of an aqueous solution we should consider -log of total [H+] in the solution. In this case Jan's explanation is very good. For quick result find out the H+ concentration from solute and then from H+ ion concentration from solvent. H+ ion concentration from water is generally less than 10^-7 due to common ion effect. $\endgroup$ – narendra kumar Jan 22 '17 at 14:28
  • $\begingroup$ @narendrakumar Yes, we can call that the condensed essence stemming from the reations between equations $(3)$ and $(4)$. $\endgroup$ – Jan Jan 22 '17 at 15:03
  • $\begingroup$ This is wrong!! If $$\ce{[H^+] =} 1.1\cdot10^{-7}$$ then $$\ce{[OH^-]} = 0.91\cdot10^{-7}$$ and $$\ce{[H^+] - [OH^-] =} 1.9\cdot10^{-8}$$. $\endgroup$ – MaxW Jan 22 '17 at 21:00
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    $\begingroup$ @MaxW Ah. You used the simplification that deprotonation is complete. Which is true for all practical intents and purposes, but I attempted to go ‘all the way’ and account for any residual $\ce{HCl}$. I believed it might not be fully irrelevant at these concentrations. $\endgroup$ – Jan Jan 22 '17 at 22:45
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    $\begingroup$ Yes, I see your point. If $\text{K}_a$ is not much greater than $\text{K}_w$ then you'll end up with a cubic equation which is a mess to solve by hand. I'd also worry about the other end. I'd doubt that 12 molar HCl is completely dissociated. $1.00\cdot10^{-8}$ HCl would be completely dissociated. $\endgroup$ – MaxW Jan 22 '17 at 22:55

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