Question

If equal volumes of HCl clout ions whose pH is 2 and 0.03 M NaOH solution are mixed;what will be the pH of the resulting solution?

Answer 1

36342, Inorganic Chemistry

If equal volumes of HCl clout ions whose pH is 2 and 0.03 M NaOH solution are mixed; what will be the pH of the resulting solution?

Solution

The pH of a solution is defined as the negative logarithm of the molar concentration of hydrogen ion:

$$\mathrm{pH} = -\lg [\mathrm{H}^{+}]$$

. If we have solution with

$$\mathrm{pH} = 2$$

, it means that we have 0.01 mole of

$$\mathrm{H}^{+}$$

per liter:

$$2 = - \lg [ H ^ {+} ]; [ H ^ {+} ] = 1 0 ^ {- 2} = 0. 0 1 \mathrm {m o l} / \mathrm {L}.$$

Also, we have 0.03 M NaOH solution, it means we have 0.03 mole of OH per liter.

Following the equation of reaction between HCl and NaOH :

$$\mathrm {H C l} + \mathrm {N a O H} = \mathrm {H} _ {2} \mathrm {O} + \mathrm {N a C l},$$

or in ionic form:

$$\mathrm {H} ^ {+} + \mathrm {O H} ^ {-} = \mathrm {H} _ {2} \mathrm {O}.$$

we have an excess of NaOH: 0.03-0.01=0.02 mol/L.

Use the ion product =

$$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] = 10^{-14}$$

to find

$$\left[\mathrm{H}^{+}\right]$$

:

$$\left[ \mathrm {H} ^ {+} \right] = \frac {\text {i o n p r o d u c t}}{\left[ \mathrm {O H} \right]} = \frac {1 \cdot 1 0 ^ {- 1 4}}{0 . 0 2} = 5 \cdot 1 0 ^ {- 1 3}.$$

Next, we can use the pH equation:

$$\mathrm {p H} = - \lg [ \mathrm {H} ^ {+} ] = - \lg (5 \cdot 1 0 ^ {- 1 3}) = 1 2. 3.$$

Answer: the pH of the resulting solution is 12.3.