Answer: The concentration of the weak acid in the sample is 0.1224 M
Explanation:
The balanced chemical reaction is :
[tex]HA+NaOH\rightarrow NaA+H_2O[/tex]
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HA[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?\\V_1=15.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 15.00=1\times 0.1036\times 17.73\\\\M_1=0.1224[/tex]
Thus the concentration (in M) of the weak acid in the sample is 0.1224
The concentration of the weak acid, HA in the sample is 0.123 M
We'll begin by calculating the number of mole of NaOH in the solution. This can be obtained as follow:
Molarity of NaOH = 0.1036 M
Volume = 17.73 mL = 17.73 / 1000 = 0.01773 L
Mole = Molarity x Volume
Mole of NaOH = 0.1036 × 0.01773
HA + NaOH —> NaA + H₂O
From the balanced equation above,
1 mole of HA reacted with 1 mole of NaOH
Therefore,
0.00184 mole of HA will also react with 0.00184 mole of NaOH.
Mole of HA = 0.00184 mole
Volume = 15 mL = 15 / 1000 = 0.015 L
Concentration = mole / Volume
Concentration of HA = 0.00184 / 0.015
Therefore, the concentration of the weak acid, HA in the sample is 0.123 M
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