Murcury is the only metal at room temperature. Its density is 13.6g/mL. How many grams of murcury will occupy a volume of 95.8mL
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Answer : The mass of sodium fluoride added should be 0.105 grams.
Explanation : Given,
The dissociation constant for HF =
Concentration of HF (weak acid)= 0.0310 M
First we have to calculate the value of .
The expression used for the calculation of is,
Now put the value of in this expression, we get:
Now we have to calculate the concentration of NaF.
Using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
Now we have to calculate the moles of NaF.
Now we have to calculate the mass of NaF.
Therefore, the mass of sodium fluoride added should be 0.105 grams.