To solve for the pH of a 0.10 M aqueous solution of a weak acid HA without ignoring xxx, you can follow these steps using an ICE table and the equilibrium expression:

Step 1: Write the dissociation equation

HA⇌H++A−HA\rightleftharpoons H^++A^-HA⇌H++A−

Step 2: Set up an ICE table for concentrations

Species| Initial (M)| Change (M)| Equilibrium (M)
---|---|---|---
HA| 0.10| −x-x−x| 0.10−x0.10-x0.10−x
H+H^+H+| 0| +x+x+x| xxx
A−A^-A−| 0| +x+x+x| xxx

Step 3: Write the equilibrium expression for the acid dissociation

constant KaK_aKa​

Ka=[H+][A−][HA]=x⋅x0.10−x=x20.10−xK_a=\frac{[H^+][A^-]}{[HA]}=\frac{x\cdot x}{0.10-x}=\frac{x^2}{0.10-x}Ka​=[HA][H+][A−]​=0.10−xx⋅x​=0.10−xx2​

Step 4: Solve for xxx

Rearrange to form a quadratic equation:

Ka(0.10−x)=x2K_a(0.10-x)=x^2Ka​(0.10−x)=x2

Ka×0.10−Kax=x2K_a\times 0.10-K_ax=x^2Ka​×0.10−Ka​x=x2

x2+Kax−0.10Ka=0x^2+K_ax-0.10K_a=0x2+Ka​x−0.10Ka​=0

Solve this quadratic equation for xxx (which represents [H+][H^+][H+]) using the quadratic formula:

x=−Ka±(Ka)2+4×0.10×Ka2x=\frac{-K_a\pm \sqrt{(K_a)^2+4\times 0.10\times K_a}}{2}x=2−Ka​±(Ka​)2+4×0.10×Ka​​​

Choose the positive root as the physically meaningful solution.

Step 5: Calculate the pH

pH=−log⁡[H+]=−log⁡xpH=-\log [H^+]=-\log xpH=−log[H+]=−logx

This method explicitly accounts for xxx in the denominator and does not use the common approximation 0.10−x≈0.100.10-x\approx 0.100.10−x≈0.10, which is valid only if xxx is very small compared to 0.10 M

Summary of the key expressions to fill in:

• ICE Table:

Species| Initial| Change| Equilibrium
---|---|---|---
HA| 0.10| −x-x−x| 0.10−x0.10-x0.10−x
H+H^+H+| 0| +x+x+x| xxx
A−A^-A−| 0| +x+x+x| xxx

• Equilibrium expression:

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

• Quadratic equation:

x2+Kax−0.10Ka=0x^2+K_ax-0.10K_a=0x2+Ka​x−0.10Ka​=0

• pH calculation:

pH=−log⁡xpH=-\log xpH=−logx

This approach ensures an accurate pH calculation for a weak acid solution without ignoring xxx.