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−Kax=x2
x2+Kax−0.10Ka=0x^2+K_ax-0.10K_a=0x2+Kax−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+]=−logxpH=-\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+Kax−0.10Ka=0
- pH calculation:
pH=−logxpH=-\log xpH=−logx
This approach ensures an accurate pH calculation for a weak acid solution without ignoring xxx.
