Question

What is the pH of a solution prepared by adding 1.59 g of sodium nitrite to 195 mL of water? Ka of HNO2 is 4.5 x 10-4.

Answer 1

solution Given 1.59 g of NaNo dissolved in 195 mL od water and formy a solution W = 1.59 g weight at Nanoz Gram molecular weight of NaNO2 volume a NaNo solution v= 195ML = m = 69 g/mol : Molarity of Nanda solution weight of Nana 1000 Gram molecular ht & NaNO2 Volume of weight solution (m) 1.59 X wro 69 g/mol 195 1590 13455 0.12 M. Concentration od Nanoz solution = c = 0.12 M

- log 4.5 - log 104 Kad HNO2 = 4.58 104 pka - log ka LO -log 4.5 x 10 10 LO - log4.5-(-4) log LO lo lo = -0.65 +4 IN 3.35 NaNO₂ is salt do weak acid and =/ ( po kwt pka + log) strong! a base pl We know that Kw = loll pk = -log ko - log loly =14 ph 12 Ź (4 +3.35 + log 0.12

pH = ² ( 14+3.35 - 11 - 0.92) - Ź (17.75 -0.92) = 16.83 2 - 8.415 pH d Nanoa solution = 8.415