Convert 1.06 g Ni(OH)2 to g NaOH. That's
1.06 x (2*molar mass NaOH/molar mass Ni(OH)2) = ?. That's the g NaOH in excess.
How many g NaOH was in the original? That must be g excess NaOH + g titrated by HCl. How much was titrated by HCl? That's L x M x molar mass NaOH = 0.090 x 0.40 x 40 = ? total g NaOH
Convert to mols NaOH, then M = mols/L.
Post your work if you get stuck.
. A student added 40.0 mL of an NaOH solution to 90.0 mL of 0.400 M HCl. The
solution was then treated with an excess of nickel(II) nitrate, resulting in the formation
of 1.06 g of Ni(OH)2 precipitate. Determine the concentration of the original NaOH
solution.
1 answer