Respuesta :
Answer:
Ca = 0.1145M
Explanation:
Volume of Acid (Va) = 80.0ml
Concentration of acid (Ca) = ?
Volume of base (Vb) = 126ml
Concentration of base (Cb) = 0.218M
The balanced chemical equation is given as;
3KOH + H3PO4 --> K3PO4 + 3H2O
The formular relaing the parameters is given as;
(CaVa) / (CbVb) = Na/Nb
Where Na = Number of moles of Acid = 1
Nb = Number of moles of Base = 3
Making Ca subject of interest;
Ca = (Na * Cb * Vb) / (Nb * Va)
Ca = (1 * 0.218 * 126) / (3 * 80)
Ca = 27.468 / 240
Ca = 0.1145M
The concentration of the H₃PO₄ solution (in M) is 0.114 M
To determine the concentration of the H₃PO₄ solution,
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
Now, from the formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B} }=\frac{n_{A} }{n_{B} }[/tex]
Where [tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
[tex]n_{B}[/tex] is the mole ratio of base
From the question
[tex]C_{B} = 0.218\ M[/tex]
[tex]V_{A} = 80.0 \ mL[/tex]
[tex]V_{B} = 126 \ mL[/tex]
From the balanced chemical equation for the reaction
[tex]n_{A} = 1[/tex]
[tex]n_{B} = 3[/tex]
Now, put all the above parameters into the equation
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B} }=\frac{n_{A} }{n_{B} }[/tex]
We get
[tex]\frac{C_{A} \times 80.0 }{0.218 \times 126}=\frac{1}{3}[/tex]
Then,
[tex]C_{A} = \frac{1 \times 0.218 \times 126 }{3 \times 80.0}[/tex]
[tex]C_{A} = \frac{27.468 }{240}[/tex]
[tex]C_{A} = 0.11445 \ M[/tex]
[tex]C_{A} \approx 0.114 \ M[/tex]
Hence, the concentration of the H₃PO₄ solution (in M) is 0.114 M
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