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
Answer: look on my page ill help you with this one but im not sure if it would be right because i dont really understand myself but look on mine
Explanation:
Answer:
Explanation:
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I suck at chemistry but i have a friend that can help
What you have to keep in mind is that
pH = - log[H+]
so 3.75 = -log[H+], we get [H+]=

,
that is [H+]≈1.77827941×

(i see this result on calculator)
this is just what you want!