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Post-Lab Questions 1. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100MCH_(3)COOH. (K_(a)=1.8times 10^-5)

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Post-Lab Questions
1. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of
0.100MCH_(3)COOH.
(K_(a)=1.8times 10^-5)

Post-Lab Questions 1. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100MCH_(3)COOH. (K_(a)=1.8times 10^-5)

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ZelieAdvanced · Tutor for 1 years

Answer

The pH of the solution is 11.72.

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## Step 1<br />First, we need to calculate the new concentrations of \(CH_3COOH\) and \(NaOH\) after the addition of the two solutions. <br /><br />### **\[ [CH_3COOH] = \frac{0.100M \times 50.0mL}{95.0mL} \approx 0.0526M \]**<br />### **\[ [NaOH] = \frac{0.100M \times 45.0mL}{95.0mL} \approx 0.0474M \]**<br /><br />## Step 2<br />With the new concentrations, we create the following initial equilibrium system once \(NaOH\) is added to the \(CH_3COOH\).<br /><br />### **\[ CH_3COOH (aq) + NaOH (aq) \rightleftharpoons CH_3COO^- (aq) + H_2O (l) \]**<br /><br />Initial:<br /><br />### **\[ [CH_3COOH] = 0.0526M, [NaOH] = 0.0474M, [CH_3COO^-] = 0, [H_2O] = 0 \]**<br /><br />Change:<br /><br />### **\[ [CH_3COOH] = -x, [NaOH] = -x, [CH_3COO^-] = x, [H_2O] = x \]**<br /><br />Final:<br /><br />### **\[ [CH_3COOH] = 0.0526M - x, [NaOH] = 0.0474M - x, [CH_3COO^-] = x, [H_2O] = x \]**<br /><br />## Step 3<br />We now plug this into the Ka expression for acetic acid.<br /><br />### **\[ 1.8 \times 10^{-5} = \frac{[CH_3COO^-][H_2O]}{[CH_3COOH]} \]**<br /><br />Substituting the values, we get:<br /><br />### **\[ 1.8 \times 10^{-5} = \frac{(0.0526M - x)(x)}{0.0526M - x} \]**<br /><br />Solving this quadratic equation, we find \(x = 0.0052M\), which represents the concentration of \(OH^-\).<br /><br />## Step 4<br />We now solve for pOH and then for pH.<br /><br />### **\[ pOH = -\log[OH^-] = 2.28 \]**<br />### **\[ pH = 14.00 - pOH = 14.00 - 2.28 = 11.72 \]**
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