Answer:
THE MILLILITERS OF 0.656 M KOH REQUIRED TO PRECIPITATE ALL THE Co2 IONS IN 187 mL OF 0.745 M Co(NO3)2 SOLUTION IS 212.37 mL
EQUATION FOR THE REACTION IS :
2 KOH + Co(NO3)2 ----------> Co(OH)2 + 2 KNO3
Step-by-step explanation:
Using dilution formula:
M1V1 = M2V2
V2 = M1 V1 / M2
M1 = 0.745 M
V1 = 187 mL
M2 = 0.656 M
V2 = unknown
V2 = 0.745 * 187 / 0.656
V2 = 139.315 / 0.656
V2 = 212.37 mL
the number of milliliters of 0.656 M KOH required to precipitate all of the Co 2 ions is 212.37 mL.
The equation for the reaction is:
2KOH + Co(NO3)2 ----------> Co(OH)2 + 2KNO3
That is 2 moles of potassium hydroxide react with 1 mole of cobalt(11) nitrate to form 1 mole of cobalt hydroxide and 2 moles of potassium nitrate