Question

\(1\ \text{L}\) of \(0.02\ \text{M}\) aqueous \(HCl\) is mixed with \(1\ \text{L}\) of \(0.01\ \text{M}\) aqueous \(H_2SO_4\) solution. Assuming complete dissociation and no change in the volume upon mixing, the pH of resultant solution is \((\log_{10}2=0.3)\):

• A. \(1.7\)
• B. \(2.7\)
• C. \(3.7\)
• D. \(2.0\)

Hint:

For strong acid mixtures, calculate total moles of \(H^+\), divide by total volume, and then use: \[ pH=-\log[H^+] \]

Answer 1

The Correct Option is A

Step 1: Calculate moles of \(H^+\) from \(HCl\).
Since \(HCl\) is a strong acid, it dissociates completely: \[ HCl \rightarrow H^+ + Cl^- \] Given, \[ M=0.02\ \text{M},\quad V=1\ \text{L} \] Moles of \(HCl\): \[ =0.02\times1=0.02 \] So, moles of \(H^+\) from \(HCl\): \[ 0.02 \]

Step 2: Calculate moles of \(H^+\) from \(H_2SO_4\).
Assuming complete dissociation: \[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \] Given, \[ M=0.01\ \text{M},\quad V=1\ \text{L} \] Moles of \(H_2SO_4\): \[ =0.01\times1=0.01 \] Moles of \(H^+\) from \(H_2SO_4\): \[ =2\times0.01=0.02 \]

Step 3: Find total concentration of \(H^+\).
Total moles of \(H^+\): \[ 0.02+0.02=0.04 \] Total volume after mixing: \[ 1+1=2\ \text{L} \] Therefore, \[ [H^+]=\frac{0.04}{2} \] \[ [H^+]=0.02\ \text{M} \]

Step 4: Calculate pH.
\[ pH=-\log[H^+] \] \[ pH=-\log(0.02) \] \[ pH=-\log(2\times10^{-2}) \] \[ pH=2-\log2 \] Given, \[ \log2=0.3 \] Therefore, \[ pH=2-0.3 \] \[ pH=1.7 \]

Step 5: Final conclusion.
Therefore, the pH of the resultant solution is \[ \boxed{1.7} \]