1. We need to identify the statement that is not true.

2. Check option A:

For aqueous solutions at $25^\circ C$, $K_w = [H^+][OH^-] = 10^{-14}$ Taking negative logarithm, $pH + pOH = 14$ So this is taken as true for aqueous solutions at $25^\circ C$.

So, A is true.

3. Check option B:

Given $1 \times 10^{-8}\,M$ HCl.

A common mistake is to say: $pH = -\log(10^{-8}) = 8$ But HCl is an acid, so the solution cannot have $pH = 8$.

Reason: at such low concentration, contribution of $H^+$ from water cannot be neglected.

Let total $[H^+] = x$. Water contributes $10^{-7}\,M$ approximately, and acid contributes $10^{-8}\,M$ additionally, so actually $[H^+] > 10^{-7}$. Hence, $pH < 7$ Certainly not 8.

More accurately, for strong acid concentration $C = 10^{-8}$, $[H^+] = \frac{C + \sqrt{C^2 + 4K_w}}{2}$ with $K_w = 10^{-14}$. Thus, $[H^+] = \frac{10^{-8} + \sqrt{10^{-16} + 4\times 10^{-14}}}{2}$ $\approx \frac{10^{-8} + 2.000\times 10^{-7}}{2}$ $\approx 1.05\times 10^{-7}$ So, $pH = -\log(1.05\times 10^{-7}) \approx 6.98$ Thus statement B is false.

So, B is not true.

4. Check option C:

By Faraday's law, 1 faraday $= 96{,}500$ C deposits 1 gram equivalent mass.

For copper in $CuSO_4$: $Cu^{2+} + 2e^- \to Cu$ Equivalent mass of copper $= \frac{\text{atomic mass}}{2}$ So 1 faraday deposits 1 gram equivalent of copper.

Thus, C is true.

5. Check option D:

Conjugate base is formed by loss of one proton: $H_2PO_4^- \to HPO_4^{2-} + H^+$ So the conjugate base of $H_2PO_4^-$ is indeed $HPO_4^{2-}$.

Thus, D is true.

6. Final conclusion:

The statement which is not true is: $\boxed{B}$