A chemistry problem by Anik Mandal

HCl is $10^{-2}$ M (Since pH=2). ie, there are $200\times 10^{-2}\times 10^{-3}$ moles of $H^+$ ions present in 300 ml HCl.

NaOH is also $10^{-2}$ M (Since pH=12). ie, there are $300\times 10^{-2}\times 10^{-3}$ moles of $OH^-$ ions present in 200 ml HCl.

When they react, NaCl is formed & a remaining of $100\times 10^{-2}\times 10^{-3}$ moles of $OH^-$ ions are present in a total volume of 500ml (= 300 mL + 200 ml).

Hence the concentration of $OH^-$ , $[OH^{-1}]=\frac{100\times 10^{-2}\times 10^{-3} moles}{500\times 10^{-3} L}=2\times10^{-3} M$

Hence pH = 14+log( $[OH^{-}]$ )=11.301

$\ce{HCl + NaOH -> Z}$

We're given that there are 200 mL of $\ce{HCl}$ and 300 mL of $\ce{NaOH}$ . In SI units:

$0.2 \text{ L}$ of $\ce{HCl}$

$0.3 \text{ L}$ of $\ce{NaOH}$

We're also given the $\ce{pH}$ of both solutions. NOTE: $\ce{pH + pOH = 14}$

$\ce{pH}_{\text{ HCl}}=2$

$\ce{pH}_{\text{ NaOH}}=12$

$\textbf{STEP 1: Calculate \# moles of HCl}$

$(i) . . .$ $\ce{ [H^+] } = 10^{-\ce{pH}}$

$(ii) . . .$ $\ce{ [H^+] } = \dfrac{ \text{ mols solute}}{ \text{ L soln}}$

From $(i)$ , we find that $\ce{ [H^+] } = 10^{-2} = 0.01 \text{ M}$ .

Knowing this molarity as well as the volume of $\ce{HCl}$ , we can use $(ii)$ to find the number of moles:

$0.01 \text{ M} = \dfrac{ n \text{ mols}}{ 0.2 \text{ L}} \Rightarrow n = 0.01 \text{ M} \times 0.2 \text{ L} = 0.002 \text{ moles } \ce{HCl}$

$\textbf{STEP 2: Calculate \# moles of NaOH}$

NOTE: $\ce{pH + pOH = 14}$

$\ce{pH = 12} \Rightarrow \ce{pOH = 2}$

From $(i)$ , we find that $\ce{ [OH^-] } = 10^{-2} = 0.01 \text{ M}$ .

Knowing this molarity as well as the volume of $\ce{NaOH}$ , we can use $(ii)$ to find the number of moles:

$0.01 \text{ M} = \dfrac{ n \text{ mols}}{ 0.3\text{ L}} \Rightarrow n = 0.01 \text{ M} \times 0.3 \text{ L} = 0.003 \text{ moles } \ce{NaOH}$

$\textbf{STEP 3: Calculate Molarity of solution Z}$

Since $0.002 \text{ mols } \ce{HCl}$ are reacting with $0.003 \text{ mols } \ce{NaOH}$ , the $0.002 \text{ mols } \ce{HCl}$ will neutralize with $0.002 \text{ mols } \ce{NaOH}$ , leaving $0.001 \text{ mols } \ce{NaOH}$ in the solution $\ce{Z}$ .

The volume of solution $\ce{Z}$ will be $0.2 \text{ L} + 0.3 \text{ L} = 0.5 \text{ L}$ .

$\ce{ [ Z ] } = \dfrac{0.001 \text{ mols}}{0.5 \text{ L}}=0.002 \text{ M}$

$\textbf{STEP 4: Find pH of Z}$

Now that we know the molarity of $\ce{Z}$ , we can calculate its $\ce{pOH}$ .

$\ce{pOH} = -log \ce{[pOH^-]} = -log(0.0020 \text{ M}) \approx 2.698$

And since $\ce{pH = 14 - pOH}$ , that means that the $\boxed{\ce{pH} \approx 11.31}$