Question

What is the pH of a 0.075 M aqueous solution of sodium alanate? (pKa1= 2.34 and pKa2 = 9.87). Report the answer to 2 places past the decimal.

Answer 1

1. PH of 0.075M aqueous solution of sodium alanate Given Data pkı, 2.34 pkaq -9.87 Formula med pka = -logka Kata² [++] = x-C pH = -log (HT) solution ) if pka, = 2.34 pka, =- log ka log ka = -pka, = -2.34 Ka= 10 234 4.57 7103 Next step ka=d²c . 4.5780 1² *0.015 6.09810 = x2 4= 2.46810 Next step [++] = x.C = 2.46810 X 0.075 [ht] = 1.85 X 10 M Next step pH = -log[++] = -log (1.85 x102) = 1.732

next is if Pka, = 9.87 H) Next -997 pka = -logka log ka aapka =nq.8T Ka= 10 = 1x10 Next soep ka=x²xC 1x10o = 22 x 0.075 -16 8²= 1x1010 0.0T5 - - = 1x1010 750103 1x107 75 1333x10 x= 3.65 x10 5 Next step = 3.65810 x 0.075. ē 2.73X100 M Next step pH = -log[ht] = -log(2.738106) = 5.56 PH, is 1.73 at Pka, = 2:34 PHz is 5.56 at pka z = 9.87.