Question

what os the ph of the solution made by mixing equal volumes of two solutions having ph=4 and ph=6 respectively.

Answer 1

What is the pH of the solution made by mixing equal volumes of solutions having $\mathsf{pH}_1 = 4$ and $\mathsf{pH}_2 = 6$ ?

Solution: As it is known, $\mathsf{pH} = -\lg [\mathsf{H}^{+}]$ , then $[\mathsf{H}^{+}] = 10^{-\mathsf{pH}}$ . After the mixing of solutions, the volume of the system doubles, and the amounts of substance of protons are adding.

$$[ H ^ {+} ] _ {1} = 1 0 ^ {- p H 1} = 1 0 ^ {- 4} M; [ H ^ {+} ] _ {2} = 1 0 ^ {- p H 2} = 1 0 ^ {- 6} M;$$ $$n \left(H ^ {+}\right) _ {1} = V \cdot \left[ H ^ {+} \right] _ {1} = V \cdot 1 0 ^ {- 4} m o l; n \left(H ^ {+}\right) _ {2} = V \cdot \left[ H ^ {+} \right] _ {2} = V \cdot 1 0 ^ {- 6} m o l;$$ $$\left[ H ^ {+} \right] _ {2} = \frac {n \left(H ^ {+}\right) _ {1} + n \left(H ^ {+}\right) _ {1}}{2 V} = \frac {V \cdot 1 0 ^ {- 4} + V \cdot 1 0 ^ {- 6}}{2 V} = 5. 0 5 \cdot 1 0 ^ {- 5} M;$$ $$\mathrm {p H} _ {1} = - \lg \left[ \mathrm {H} ^ {+} \right] _ {1} = - \lg (5. 0 5 \cdot 1 0 ^ {- 5}) = 4. 3.$$

Answer: 4.3.