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if 10.57 mL of an acetic Acid required 10.84 mL of 0.9221 M NaOH to reach...

if 10.57 mL of an acetic Acid required 10.84 mL of 0.9221 M NaOH to reach the endpoint, what’s the concentration of Acetic Acid solution?
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Answer #1

given volume of acetic acid = V1=10.57 mL

concentration of acetic acid = Say= M1

concentration of NaOH =0.9221

volume of NaOH = 10.84 mL

now no. of moles of acetic acid = no. of moles of NaOH

M1 *V1 =M2*V2

M1*10.57 = 0.9221 *10.84

M1 = 0.9457M

thus concentration of acetic acid = 0.9457M

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