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Sagot :
Answer:
[tex]C:\text{ 0.08 M}[/tex]Explanation:
Here, we want to get the molarity of the base
To get this, we start by writing the balanced equation of reaction between the base and the acid
We have that as:
[tex]Ca(OH)\placeholder{⬚}_{2(aq)}\text{ + 2HC}_2H_3O_{2(aq)}\text{ }\rightarrow\text{ Ca\lparen CH}_3COO)\placeholder{⬚}_{2(aq)}\text{ + 2H}_2O_{(l)}[/tex]Now, we proceed to write the standardization equation
We have that as:
[tex]\frac{C_aV_a}{C_bV_b}\text{ = }\frac{n_a}{n_b}[/tex]where:
Ca is the molarity of the acid which is 0.1M
Va is the volume of the acid which is 31.5 mL
Cb is the molarity of the base which is what we want to calculate
Vb is the volume of the base which is 19.4 mL
na is the number of moles of the acid in the balanced equation which is 2
nb is the number of moles of the base which is 1 in the balanced equation
Substituting the values, we have:
[tex]\begin{gathered} \frac{0.1\text{ }\times\text{ 31.5}}{C_b\times\text{ 19.4}}\text{ = }\frac{2}{1} \\ \\ C_b\text{ = }\frac{31.5\text{ }\times\text{ 0.1}}{2\times19.4}\text{ = 0.08 M} \end{gathered}[/tex]
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