Question

4ml of hcl solution of ph=2 is mixed with 6ml of naoh solution of ph=12. What would be the final ph of solution?

Answer 1

It becomes neutral and final pH becomes 7

Answer 2

Answer: The final pH of the solution is 10.30

Explanation:

We are given:

Volume of HCl = 4 mL

Volume of NaOH = 6 mL

• To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 2

Putting values in above equation, we get:

$2=-\log[H^+]$

$[H^+]=10^{-2}$

• To calculate the hydroxide ion concentration, we first calculate pOh of the solution, which is:

pH + pOH = 14         ........(1)

We are given:

pH of NaOH = 12

$pOH=14-12=2$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$            .......(2)

We are given:

pOH = 2

Putting values in above equation, we get:

$2=-\log[OH^-]$

$[OH^-]=10^{-2}$

• To calculate the milli equivalents, we use the equation:

$M_{eq}=\text{Molarity}\times \text{Volume (in mL)}$

The chemical equation for the reaction of HCl and NaOH follows:

                                     $HCl+NaOH\rightarrow NaCl+H_2O$

Meq before reaction:   0.004   0.006         0     0

Meq after reaction:      0        0.002       0.004    0.004

Concentration of hydroxide ions after the reaction:

$[OH^-]=\frac{0.002}{10}=2\times 10^{-4}M$

Calculating the pOH by using equation 2, we get:

$pOH=-\log(2\times 10^{-4})\\\\pOH=3.70$

Now, calculating the value of pH by using equation 1, we get:

$pH+3.70=14\\\\pH=10.30$

Hence, the final pH of the solution is 10.30