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When NH(3)(0.1 M) 50 ml mix with HCl (0....

When `NH_(3)(0.1 M) 50 ml` mix with `HCl (0.1 M) 10 ml` then what is `pH` of resultant solution (`pK_(b)=4.75)`

A

a. 9.25

B

b. 10

C

c. 9.85

D

d. 4.15

Text Solution

AI Generated Solution

The correct Answer is:
To find the pH of the resultant solution when 50 mL of 0.1 M NH3 is mixed with 10 mL of 0.1 M HCl, we can follow these steps: ### Step 1: Calculate the moles of NH3 and HCl - **Moles of NH3**: \[ \text{Moles of NH3} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} = 5 \, \text{mmol} \] - **Moles of HCl**: \[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.010 \, \text{L} = 0.001 \, \text{mol} = 1 \, \text{mmol} \] ### Step 2: Determine the reaction and the moles after the reaction The reaction between NH3 and HCl can be represented as: \[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^- \] - **Initial moles**: - NH3: 5 mmol - HCl: 1 mmol - NH4Cl (produced): 0 mmol - **Final moles after reaction**: - NH3: \(5 - 1 = 4 \, \text{mmol}\) - HCl: \(1 - 1 = 0 \, \text{mmol}\) - NH4Cl: \(0 + 1 = 1 \, \text{mmol}\) ### Step 3: Calculate the concentrations of NH3 and NH4Cl in the final solution - **Total volume of the solution**: \[ 50 \, \text{mL} + 10 \, \text{mL} = 60 \, \text{mL} = 0.060 \, \text{L} \] - **Concentration of NH3**: \[ [\text{NH}_3] = \frac{4 \, \text{mmol}}{60 \, \text{mL}} = \frac{4 \times 10^{-3} \, \text{mol}}{0.060 \, \text{L}} = 0.0667 \, \text{M} \] - **Concentration of NH4Cl**: \[ [\text{NH}_4^+] = \frac{1 \, \text{mmol}}{60 \, \text{mL}} = \frac{1 \times 10^{-3} \, \text{mol}}{0.060 \, \text{L}} = 0.0167 \, \text{M} \] ### Step 4: Calculate the pOH using the Henderson-Hasselbalch equation Using the formula: \[ \text{pOH} = \text{pK}_a + \log\left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right) \] First, we need to find pKa from pKb: \[ \text{pK}_a + \text{pK}_b = 14 \implies \text{pK}_a = 14 - 4.75 = 9.25 \] Now substituting the values: \[ \text{pOH} = 9.25 + \log\left(\frac{0.0167}{0.0667}\right) \] Calculating the log term: \[ \log\left(\frac{0.0167}{0.0667}\right) \approx \log(0.25) \approx -0.602 \] Thus: \[ \text{pOH} = 9.25 - 0.602 \approx 8.648 \] ### Step 5: Calculate the pH Using the relation: \[ \text{pH} + \text{pOH} = 14 \] We find: \[ \text{pH} = 14 - 8.648 \approx 5.352 \] ### Final Answer The pH of the resultant solution is approximately **5.35**.
To find the pH of the resultant solution when 50 mL of 0.1 M NH3 is mixed with 10 mL of 0.1 M HCl, we can follow these steps: ### Step 1: Calculate the moles of NH3 and HCl - **Moles of NH3**: \[ \text{Moles of NH3} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol} = 5 \, \text{mmol} \] ...
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