Step 1: Plan:
Treat it as the equilibrium $\text{CH}_3\text{CH}_2\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{CH}_2\text{COO}^-$ with the salt supplying the common ion.
Step 2: Use the equilibrium expression:
\[ [\text{H}^+] = K_a\frac{[\text{acid}]}{[\text{salt}]} \]
The two solutions have the same molarity and the same volume, so the ratio of acid to salt is $1$. Therefore $[\text{H}^+] = K_a = 1.3\times10^{-5}$ M.
Step 3: Convert to pH:
\[ \text{pH} = -\log(1.3\times10^{-5}) = 4.89 \]
A value of $2.45$ would need a strong acid. Values near $6$ to $7$ would mean little acid is left, which is not the case here.
Final Answer:
The pH is $4.89$, so option (B) is correct.
\[ \boxed{4.89} \]