10 mL of 0.1 M NaOH is equal to 1 equivalent of base, if titrating 10.00 mL of 0.1 M amino acid.
Given :
0.1 M NaOH
0.1 M Amino acid
10.00 mL volume of Amino acid
<em> R-CH(NH₂)-COOH + NaOH → R-CHNH₂Na + H₂O</em>
In this case, end factor = 1 and It is 1 mole / 1 mole ratio
Since, the concentration of amino acid and NaOH are same, that is 0.1 M.
Concentration of amino acid = concentration of NaOH
0.1 M = 0.1 M
Therefore, 10 mL of NaOH is required to neutralize 10 mL of amino acid and the end factor is equivalent to 1.
Hence, The volume of 0.1 M NaOH is equal to 1 equivalent of base.
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