If 25 mls of 0.01 M NaOH is added to 75 mls of 0.005 HCl, what is the resulting pH

pH of the resulting solution will be 2.9.

Explanation:

25 ml of 0.01 M NaOH contains = 25 × 0.01 = 0.25 milli moles

75 ml of 0.005 M HCl contains = 75 × 0.005 = 0.375 milli moles

(0.375 - 0.25) = 0.125 milli moles

0.125 milli moles of each NaOH and HCl may neutralize each other and so the final solution may contain 0.125 milli moles of HCl.

In 1 L or 1000 ml of the solution, the quantity of HCl present is 0.125 × 10 = 1.25 milli moles

It can be converted to moles by dividing milli moles by 1000, we will get,

0.00125 moles

So the concentration of Hydrogen ions, [H⁺] = 0.00125 M

pH = -log₁₀[H⁺]

    = -log₁₀[0.00125]

   = 2.9

So pH of the resulting solution will be 2.9.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2