Question

If you used 4.5 ml of 6.0 M HNO3 to dissolve the sample of Cu (0.350g)

What volume of 4.0M NaOH would be required to neturalized the excess HNO3

Answer 1

Cu(s) + 4 HNO3(aq) --------------> Cu(NO3)2(aq)+2NO2(g)+2H20(l)

a) Number of moles of Cu = 0.350/63.546 = 0.00551 moles

Since 1 mole of Cu reacts with 4 moles of HNO3,

Number of moles of HNO3 (required) = 4 * 0.00551 moles = 0.02204 moles

Molar mass of HNO3 = 1 + 14 + 3*16 = 63 gm/mol

Hence mass required = number of moles * molar mass = 0.02204 moles * 63 gm/mol = 1.38852 gms

b) Number of moles = 0.02204 moles

0.02204 = V/1000 * 6

Hence volume = (0.02204 * 1000)/6 = 3.673 ml

c) If we used 4.5 ml of 6M HNO3

volume of HNO3 left after the reaction = 4.5 ml - 3.673 ml = 0.827 ml

Rxn between HNO3 and NaOH

NaOH + HNO3 -----> NaNO3 + H20

Using the formula M1V1 = M2V2

0.827 x 6 = 4 x V2

V2 = 1.24 ml

volume of NaOH = 1.24 mL