Question

What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide? b.) If you added 35.0 mL of hydrochloric acid solution to the sodium hydroxide, will the resulting solution be acidic or basic? Explain your reasoning. c.) Describe how you would neutralize the new solution using amounts (in volume) of either 0.496 M HCl or 0.809 M sodium hydroxide.

Answer 1

cid (Hcl)
molarity M1 = 0.496M
volume v1    =
no of moles = 1
Base(NaOH)
molarity M2 = 0.809M
volume V2 = 20ml
no of moles = 1
M1V1/n1 = M2V2/n2
V1      = M2V2*n1/n2M1
        = 0.809*20*1/0.496*1
volume V1 = 32.62ml
volume of HCl is 32.62>>> answer
b.no of moles of HCl = molarity * volumein liters
                     = 0.496*0.035
                     = 0.01736 moles
no of moles of NaOH = molarity * volume in liters
                     = 0.809*0.02
                     = 0.01618 moles
no of moles of HCl is greater than no of moles of NaOH
resultant solution is acidic nature
c. HCl
molarity M1   = 0.496M
volume V1     = 35ml
no of moles n1= 1 moles
NaOH
molarity M2   = 0.809M
volume   v2   =
no of moles   = 1
M1V1/n1      = M2V2/n2
V2           = M1V1*n2/M2n1
             = 0.496*35*1/0.809*1
             = 21.45ML
VOLUME OF NaOH is 21.45ml