For the nitrous acid, HNO2, Ka= 4.0 x 10^-4. Calculate the ph of 0.27 M HNO2
The Correct Answer and Explanation is:
To calculate the pH of a 0.27 M solution of nitrous acid (HNO2), we need to use the acid dissociation constant (Ka) for HNO2, which is given as 4.0×10−44.0 \times 10^{-4}4.0×10−4. Here’s how we can approach the problem:
Step 1: Write the dissociation equation
The dissociation of nitrous acid in water can be written as:HNO2(aq)⇌H+(aq)+NO2−(aq)\text{HNO}_2 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{NO}_2^- (aq)HNO2(aq)⇌H+(aq)+NO2−(aq)
The equilibrium expression for the dissociation is:Ka=[H+][NO2−][HNO2]K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]}Ka=[HNO2][H+][NO2−]
Given Ka=4.0×10−4K_a = 4.0 \times 10^{-4}Ka=4.0×10−4, we will use the following assumptions:
- The concentration of HNO2\text{HNO}_2HNO2 initially is 0.27 M.
- The concentration of H+\text{H}^+H+ and NO2−\text{NO}_2^-NO2− will be the same at equilibrium because they are produced in equal amounts.
Step 2: Set up the equilibrium concentrations
Let the concentration of H+\text{H}^+H+ (and NO2−\text{NO}_2^-NO2−) at equilibrium be xxx. Initially, the concentration of HNO2\text{HNO}_2HNO2 is 0.27 M, and it decreases by xxx due to dissociation. So, at equilibrium:
- [HNO2]=0.27−x[\text{HNO}_2] = 0.27 – x[HNO2]=0.27−x
- [H+]=x[\text{H}^+] = x[H+]=x
- [NO2−]=x[\text{NO}_2^-] = x[NO2−]=x
Step 3: Apply the equilibrium expression
Substitute the equilibrium concentrations into the expression for KaK_aKa:4.0×10−4=x20.27−x4.0 \times 10^{-4} = \frac{x^2}{0.27 – x}4.0×10−4=0.27−xx2
Assume that xxx is much smaller than 0.27 (since KaK_aKa is relatively small), so we approximate 0.27−x≈0.270.27 – x \approx 0.270.27−x≈0.27. This simplifies the equation to:4.0×10−4=x20.274.0 \times 10^{-4} = \frac{x^2}{0.27}4.0×10−4=0.27×2
Step 4: Solve for xxx
Now, solve for xxx:x2=(4.0×10−4)(0.27)x^2 = (4.0 \times 10^{-4})(0.27)x2=(4.0×10−4)(0.27)x2=1.08×10−4x^2 = 1.08 \times 10^{-4}x2=1.08×10−4x=1.08×10−4=0.0104 Mx = \sqrt{1.08 \times 10^{-4}} = 0.0104 \, \text{M}x=1.08×10−4=0.0104M
Step 5: Calculate the pH
The concentration of H+\text{H}^+H+ is x=0.0104 Mx = 0.0104 \, \text{M}x=0.0104M. The pH is given by:pH=−log[H+]\text{pH} = -\log[\text{H}^+]pH=−log[H+]pH=−log(0.0104)=1.98\text{pH} = -\log(0.0104) = 1.98pH=−log(0.0104)=1.98
Conclusion
The pH of the 0.27 M HNO2 solution is approximately 1.98.
This calculation assumes that the dissociation is relatively weak, which is true for many weak acids like HNO2. The Ka value indicates that only a small fraction of the acid dissociates, so the pH is relatively low but not as acidic as a strong acid like HCl, for example.
