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