Here’s how you would fill in the blanks and set up the work to solve for the pH of a 0.10 M weak acid HA without ignoring x.

1. Write the dissociation equation

For a generic weak acid HA in water:

HA(aq)⇌H+(aq)+A−(aq)\text{HA(aq)}\rightleftharpoons \text{H}^+\text{(aq)}+\text{A}^-\text{(aq)}HA(aq)⇌H+(aq)+A−(aq)

(If your course uses hydronium, it would be HA+H2O⇌H3O++A−\text{HA}+\text{H}_2\text{O}\rightleftharpoons \text{H}_3\text{O}^++\text{A}^-HA+H2​O⇌H3​O++A−.)

2. ICE table (do not ignore x)

Let the initial concentration of HA be 0.10 M, and let xxx be the amount that dissociates at equilibrium.

Return this as HTML (as requested):

html

<table border="1" cellpadding="4" cellspacing="0">
  <tr>
    <th></th>
    <th>HA(aq)</th>
    <th>H<sup>+</sup>(aq)</th>
    <th>A<sup>-</sup>(aq)</th>
  </tr>
  <tr>
    <td>Initial (M)</td>
    <td>0.10</td>
    <td>0</td>
    <td>0</td>
  </tr>
  <tr>
    <td>Change (M)</td>
    <td>-x</td>
    <td>+x</td>
    <td>+x</td>
  </tr>
  <tr>
    <td>Equilibrium (M)</td>
    <td>0.10 - x</td>
    <td>x</td>
    <td>x</td>
  </tr>
</table>

Here, the key blanks to fill are: 0.10, 0, 0, −x, +x, +x, 0.10 − x, x, x.

3. Equilibrium (Ka) expression with x kept

The acid dissociation constant expression for HA is:

Ka=[H+][A−][HA]K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka​=[HA][H+][A−]​

Substitute the equilibrium concentrations from the ICE table (and do not approximate away x):

Ka=x⋅x0.10−x=x20.10−xK_a=\frac{x\cdot x}{0.10-x}=\frac{x^2}{0.10-x}Ka​=0.10−xx⋅x​=0.10−xx2​

So the main “fill in the blanks” here are: x , x , 0.10 − x , and the final form K a=x20.10−xK_a=\dfrac{x^2}{0.10-x}Ka​=0.10−xx2​.

4. Solving for pH (conceptual)

To actually get the pH, you would:

  1. Use the given KaK_aKa​ for HA (from your problem statement) in

Ka=x20.10−xK_a=\frac{x^2}{0.10-x}Ka​=0.10−xx2​

  1. Solve this equation for xxx (this is the equilibrium [H+][\text{H}^+][H+]).
  2. Compute

pH=−log⁡([H+])=−log⁡(x)\text{pH}=-\log([\text{H}^+])=-\log(x)pH=−log([H+])=−log(x)

In many textbook examples, solving that equation leads to a quadratic in xxx when you refuse to ignore x , which is exactly what your question is asking you to show.

Quick story-style check

Imagine you start with a “bucket” of 0.10 M HA. At equilibrium, a small chunk x leaves the bucket to become H+\text{H}^+H+ and A−\text{A}^-A−, so both of those new “buckets” have x , while the original bucket is left with 0.10 − x. The Ka expression just ties those three buckets together in the fraction x20.10−x\dfrac{x^2}{0.10-x}0.10−xx2​.

TL;DR :

  • Equation: HA⇌H++A−\text{HA}\rightleftharpoons \text{H}^++\text{A}^-HA⇌H++A−.
  • ICE table equilibrium line: [HA]=0.10−x[\text{HA}]=0.10-x[HA]=0.10−x, [H+]=x[\text{H}^+]=x[H+]=x, [A−]=x[\text{A}^-]=x[A−]=x.
  • Ka expression (no approximation): Ka=x20.10−xK_a=\dfrac{x^2}{0.10-x}Ka​=0.10−xx2​.