Question 10 of 10 The triprotic acid H, A has ionization constants of Ka141 0- 6.38 x 10-6, and K...
The triprotic acid H, A has ionization constants of Kal = 5.82 x 10-2, K:2 = 5.15 x 10-7, and K3 = 4.14 x 10-13. Calculate the Ht concentration and the ratio- [HA] for a 0.0460 M solution of NaH, A. [H2A] [H+] = [HA] [HA] Calculate the H+ concentration and the ratio ha for a 0.0460 M solution of Na, HA. "° [HA] [HP] = [HA2-) [H,A-]
The triprotic acid H3A has ionization constants of Ka1=5.37×10−3 , Ka2=4.31×10−6 , and Ka3=1.04×10−11 . Calculate the H+ concentration and the ratio [H2A−]/[H3A] for a 0.0170 M solution of NaH2A . [H+]= [H2A−]/[H3A]= Calculate the H+ concentration and the ratio [HA2−]/[H2A−] for a 0.0170 M solution of Na2HA . [ H + ]= [HA2−]/[H2A−]=
The triprotic acid H3AH3A has ionization constants of ?a1=3.38×10−3Ka1=3.38×10−3, ?a2=7.27×10−8Ka2=7.27×10−8, and ?a3=4.17×10−11Ka3=4.17×10−11. Calculate the H+H+ concentration and the ratio [H2A−][H3A][H2A−][H3A] for a 0.06400.0640 M solution of NaH2ANaH2A. [H+]=[H+]= MM [H2A−][H3A]=[H2A−][H3A]= Calculate the H+H+ concentration and the ratio [HA2−][H2A−][HA2−][H2A−] for a 0.06400.0640 M solution of Na2HANa2HA. [H+]=[H+]= MM [HA2−][H2A−]=[HA2−][H2A−]=
The triprotic acid H3AH3A has ionization constants of ?a1=3.38×10−3Ka1=3.38×10−3, ?a2=7.27×10−8Ka2=7.27×10−8, and ?a3=4.17×10−11Ka3=4.17×10−11. Calculate the H+H+ concentration and the ratio [H2A−][H3A][H2A−][H3A] for a 0.06400.0640 M solution of NaH2ANaH2A. [H+]=[H+]= MM [H2A−][H3A]=[H2A−][H3A]= Calculate the H+H+ concentration and the ratio [HA2−][H2A−][HA2−][H2A−] for a 0.06400.0640 M solution of Na2HANa2HA. [H+]=[H+]= MM [HA2−][H2A−]=[HA2−][H2A−]=
-The triprotic acid H3A has ionization constants of Ka1 = 4.9× 10–2, Ka2 = 6.0× 10–6, and Ka3 = 2.0× 10–13. Calculate the following values for a 0.0700 M solution of NaH2A (H+)=? (H2A-)/(H3A)=? -Calculate the following values for a 0.0700 M solution of Na2HA. (H+)=? (HA2-)/(H2A-)=?
A diprotic acid, H2A, has acid dissociation constants of Kai = 3.52 x 10-4 and Ka2 = 2.03 × 10-11 . Calculate the pH and molar concentrations of H2A, HA, and A2- at equilibrium for each of the solutions. A 0.206 M solution of H,A. pH = H2A] HA1 A 0.206 M solution of NaHA pH- [H2A] = [HA-] = A 0.206 M solution of Na,A. pH- [H2A] EA T [A21
A diprotic acid, H,A, has acid dissociation constants of Ka molar concentrations of H,A, HA-, and A2- at equilibrium for each of the solutions 1.42 x 10-4 and Ka2 = 4.07 x 1012. Calculate the pH and = A 0.210 M solution of H,A H2A] = pH HA- A2- М М A 0.210 M solution of NaHA HA pH= М
A diprotic acid, H2A, has acid dissociation constants of
Ka1=1.01×10−4 and Ka2=4.08×10−12. Calculate the pH and molar
concentrations of H2A, HA−, and A2−at equilibrium for each of the
solutions.
A diprotic acid, H, A, has acid dissociation constants of Kal = 1.01 x 104 and K22 = 4.08 x 10-12. Calculate the pH and molar concentrations of H, A, HA, and A? at equilibrium for each of the solutions. A 0.176 M solution of H, A. pH= pH = 1...
A diprotic acid, H,A, has acid dissociation constants of Ka1 = 2.09 x 104 and Ka2 = 3.96 x 10-11. Calculate the pH and molar concentrations of H,A, HA-, and A2- at equilibrium for each of the solutions. A 0.183 M solution of H,A pH H,A= A2-1 HA] = A 0.183 M solution of N2HA. HA pH= HA A2- A 0.183 M solution of Na, A H,A ] pH= HA A2-1 M M
A diprotic acid, H,A, has acid dissociation...
The ionization constants for the diprotic acid H 2S are 1.0 x 10 -7 and 1.3 x 10 -13. H2S (aq) + H2O (l) <=> HS -(aq) + H3O +(aq) Ka = 1.0 x 10 -7 HS -(aq) + H2O (l) <=> S-2 (aq) + H3O +(aq) Ka = 1.3 x 10 -13 What is the equilibrium concentration of HS - in a 0.300 M solution of H2S?