Answer:
278 ml of KOH.
Explanation:
The equation for the reaction is: Pb(NO3)2(aq) + 2KOH(aq) --> Pb(OH)2(s) + 2KNO3(aq)
Pb(NO3)2:
Half ionic equation:
Pb(NO3)2(aq) --> Pb2+ + 2NO3^-
Pb(OH)2 --> Pb2+ + 2OH-
Volume = 135 mL
Molar concentration = 0.775 M
Number of moles = molar concentration * volume
= 0.775 * 0.135
= 0.105 mol of Pb(NO3)2
Since 1 mole of Pb(NO3)2 reacted with 2 moles of KOH to give 1 mole of Pb2+ (Pb(OH)2).
By stoichiometry,
Number of moles of KOH = 0.105 * 2
= 0.21 mol
Molar concentration = number of moles/volume
Volume = 0.21/0.753
= 0.278 l
To ml, 0.278 l * 1000 ml/1l
= 278 ml of KOH.