First, we need to get the concentration of [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M
then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M
and when Pka of the 2nd ionization of phosphoric acid = 7.21
So by substitution in the following formula, we can get the PH:
PH = Pka + ㏒[A]/[AH]
∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2
She I study I like to read the question while covering the answer and then answer it I. My head. Then I check to see if it is right. Also if you want flash cards just search up the topic with quizlet for online quizzes about your subject. If needed take 25 mins to study with 5 min breaks.
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Answer : The mass of sodium sulfate needed is 5.7085 grams.
Explanation : Given,
Concentration of sodium ion = 0.148 mol/L
Volume of solution = 2.29 L
Molar mass of sodium sulfate = 142 g/mole
First we have to determine the moles of sodium ion.


Now we have to calculate the moles of sodium sulfate.
The balanced chemical reaction will be,

As, 2 moles of sodium ion produced from 1 moles of 
So, 0.08035 moles of sodium ion produced from
moles of 
Now we have to calculate the mass of sodium sulfate.


Therefore, the mass of sodium sulfate needed is 5.7085 grams.