(a) What is the pHof$\mathbf{0}.\mathbf{0111}\mathbf{M}\mathbf{NaOH}$? Is the solution neutral, acidic, or basic? (b) What is the pOHof$\mathbf{1}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathit{M}\mathbf{}\mathit{H}\mathit{C}\mathit{l}$? Is the solution neutral, acidic, or basic?

The negative logarithm of ${\mathbf{\text{H}}}^{\mathbf{\text{+}}}$ion concentration is used to calculate pH. As a result, the meaning of pHis justified as hydrogen power.

Let's start with the strong base's dissociation reaction. In water,$\text{NaOH}$

$\text{NaOH}\to {\text{Na}}^{\text{+}}{\text{+ OH}}^{\text{-}}$

The concentration of$\text{NaOH}$is equal to the sum of the concentrations of$\left[{\text{Na}}^{\text{+}}\right]$and$\left[{\text{OH}}^{\text{-}}\right]$.

We can solve for the pOH using this information.

$$\begin{gathered} \text{pOH = - log}\left[{\text{OH}}^{\text{-}}\right] \\ \text{= - log(0.0111 M)} \\ \text{pOH = 1.95} \end{gathered}$$

We can solve it using the equation below because we're seeking for its pH.

$$\begin{gathered} \text{pOH + pH = 14} \\ \text{pH = 14 - pOH} \\ \text{= 14 - 1.95} \\ \text{pH = 12.05} \end{gathered}$$

The solution is simple because pH>7 .

Therefore, Solution is pH=12.05 , basic.

The number of moles is –

$1.35\times {10}^{-3}MHCl$

First, let's look at how the strong acid HCl dissociates in water.

$\text{HCl}\to {\text{H}}^{\text{+}}{\text{+ Cl}}^{\text{-}}$

We can solve for the pH using this information.

$$\begin{gathered} \text{pH = - log}\left[{\text{H}}^{\text{+}}\right] \\ \text{=-log}\left(1.35\times {10}^{-3}M\right) \\ pH=2.87 \end{gathered}$$

Since we're seeking for its OH, we'll use the equation below to solve it.

$$\begin{gathered} \text{pOH + pH = 14} \\ \text{pOH = 14 - pH} \\ \text{= 14 - 2.87} \\ \text{pOH = 11.13} \end{gathered}$$

The solution is acidic because of the pH<7.

Therefore, Solution pOH = 11.13 is , acidic.