Question

can u show every step please.

CHEM 1312 Homework #16 Upload a .pdf file to Blackboard by 11:59pm on Monday, 3/30/2020 (5 points possible) Barbituric acid, HC.H.N.Ois a weak acid that is used to prepare various barbiturate drugs (used as sedatives). The value of K, at 25°C is 9.8 x 10-5 (a) pt) In the space below, write the balanced chemical equation for the equilibrium established when HC.H.N.O, is dissolved in water. () pts) Calculate the pH of a solution that is 0.20 Min HC.H3N203. Assume that the temperature is 25°C

Answer 1

. 23 By definition, Aud is a substance which gives it ion ' When dissolved in water. HCG H B No is a barbituric acid. When it is dissolved in Water at gives it ion. Equation is : HC4Hz N2Oz Lag) +420 = CHINO Lag) + H307an) 7 ph is give by : PH = -log[H30t ] 0 So, we have to find concentration of hot at ICE table H Cu H₃N2 Oz the 2 cu H ₂N2O3 + Hot Let us suppore at Initial 0.20 C o o equiliboium ze amount conce - Change -x te the l of H₃0t and CuM NOT formed from 0.204 of Equilibrium 0.20-1 x a r HC 4 H ₂ N2O3 Equilibrium constant Ka is given by : : Remaining concentration - Ka= [cu M₂ M2O3] [Hz ot] of K CG H NO at equilibrium HC4H3N2O3] I NOTE Conce, of solids = & liquide 9.8x105 - Xike is taken 0,20-30 l as unidy or 9.8x10 106)] =1 er or 72 = 1.96 xboʻ– 9.8 x10 %

me to get + 9.8*10*4 - 1960 solve this quadratic by using formula: -bt bayac. x = 2 a – 9.8x10°£ = – 4.8x109+ Van 2 = 9. Xx165)2 - 4X1X (-1.96x105) 2X1 X = ic -9.8x105 + 9.88 lūlo +7.84*1ös 2 - 9.8x105 + N0.000098x175+ 7.84x765 m2 0.000098x105 + 7.84x305 = (0.000098 +7.84) 16 5 7.84x1ö 5 l -9.8 X10 I 7 7.84x105 r = -9.8710 178.4166 or x = -9,8X105+ 8.85x103 _ -9.8x165 + 885x18 5 se = 875-2x105 -9.8x10 5 – 885x105 -9.88162 + 885x135 X= x = 2 2 From here we will get are value of x. But value of he can't be negative this can be neglected 2= 437.6X10S. 10 wound off to 4.4X10² or x = 4.4x76?] 2

At equilibriun on At equilibrium conce of thot, [4_ot] = x :[130*) = 4.4x173 M Substitute in eq" 0 PH = -log [ 4.4 x 103 [logax b= log at log b] Pre=-{log 1.44 + log103 ] [:logab = blog a] bu = - [ 0.64 - 3 log10 ] [log 4.4 = 0.64] Pu=-{0.64-3] bw=-[-2.36) = 2.36 Ampte 2:36