To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
In finding the molarity of a solution, we use the following formula:

What is Molarity?
The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.
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The molar mass of NH4NO3 and (NH4)3PO4 are 80.043 g/mol and 149.0867 g/mol, respectively.




![[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4](https://tex.z-dn.net/?f=%5BNH%2B4%5D%3D0.1596%20mol20.0%20L%3D7.98%C3%9710%E2%88%923%20M%20NH%2B4)
![[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4](https://tex.z-dn.net/?f=%5BPO3%E2%88%924%5D%3D0.0296%20mol20.0%20L%3D1.48%C3%9710%E2%88%923%20M%20PO3%E2%88%924)
Therefore,
has a molarity of 
To learn more about Molarity click on the link below:
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To solve this problem,
we can use the Henderson-Hasselbalch Equation which relates the pH to the measure
of acidity pKa. The equation is given as:<span>
<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>
Where,
[base] = concentration
of C2H3O2
in molarity or moles
<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>
For the sake of easy calculation, let us assume that:
[base] = 1
[acid] = x
<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)
<span>log (1 / x) = - 0.5
1 / x = 0.6065 </span></span>
x =
1.65<span>
The required ratio of C2H3O2 /HC2H3O2 <span>
is 1:1.65 or 3:5. </span></span>