Chemistry: Determine the pH of a solution created by mixing 95.0 mL of 0.200 M nitric acid, HNO3, with 320.0 mL of 0.078 M potassium hydroxide, KOH.

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Determine the pH of a solution created by mixing

Ответ:

Wxaieyu

30.01.2022

Chemistry

12.16

Explanation:

We have to start with the reaction between nitric acid and potassium hydroxide:

HNO_3~+~KOH~-~H_2O~+~KNO_3

With this in mind, we can calculate the number of moles of each compound if we use the molarity equation: $M=\frac{mol}{L}$ (this means that we have to find the Litters dividing by 1000)

95~mL~=~0.095~L

320~mL~=~0.32~L

Now, we can calculate the moles:

moles~=~0.2*0.095~=~0.019~mol~HNO3

moles~=~0.078*0.32~=~0.025~mol~KOH

If we have a molar ratio 1:1 (1 mol of KOH reacts with 1 mol of HNO_3 ). So, if we have 0.019~mol~HNO3 we will need 0.019~mol~KOH . So, we have to calculate the amount of KOH that remains in the solution, so:

0.025-0.019~=~0.006~mol~KOH

If we have an excess of KOH, this compound will be the cause of the pH value. Therefore we have to calculate the concentration. We already know the moles we have to calculate the volume. The total volume:

Vt~=~95mL~+~320mL~=~415~mL~=~0.415~L