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Salt hydrolysis: pH of salts of weak acid or weak base

37 questions on Salt hydrolysis: pH of salts of weak acid or weak base from the Ionic Equilibrium II chapter: 8 NEET level, 15 JEE Main, 14 JEE Advanced. Each one opens with its answer and solution.

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Which of the following salts gives an acidic aqueous solution at 298 K? NEET level
An aqueous solution of sodium cyanide, NaCN, has a pH greater than 7 because NEET level
For a salt BA of a weak base BOH (with K_b) and a strong acid, the hydrolysis constant is NEET level
Ammonium acetate is dissolved in water at 298 K. Given K_a(CH3COOH) = K_b(NH3), the solution is NEET level
Calculate the pH of a 0.10 M solution of sodium acetate. K_a(CH3COOH) = 1.8x10^-5 at 298 K. Use pH = 7 + tfrac12(pK_a + log c). JEE Main level
What is the pH of a 0.20 M solution of NH4Cl at 298 K? K_b(NH3) = 1.8x10^-5. Use pH = 7 - tfrac12(pK_b + log c). JEE Main level
The degree of hydrolysis of 0.010 M sodium fluoride at 298 K is closest to (K_a of HF = 6.8x10^-4; use h = sqrtK_h/c) JEE Main level
Ammonium formate is the salt of HCOOH (pK_a = 3.75) and NH3 (pK_b = 4.75). Use pH = 7 + tfrac12(pK_a - pK_b). The pH of its aqueous solution is JEE Main level
A 0.10 M solution of the sodium salt NaA of a weak monoprotic acid HA has pH 9.00 at 298 K. The value of K_a of HA is JEE Main level
Calculate the pH of a 0.040 M aqueous solution of sodium cyanide at 298 K. K_a(HCN) = 4.9x10^-10, K_w = 1.0x10^-14. Neglect any correction to the square-root fo... JEE Main level
The salt BA is formed from a weak acid HA (K_a = 2.0x10^-5) and a weak base BOH (K_b = 1.6x10^-5). Use K_h = K_w/(K_a K_b) and h = sqrtK_h (valid when h ll 1). ... JEE Main level
Which of the following aqueous solutions (0.1 M, 298 K) have a pH greater than 7? Given: K_a(HNO2) = 4.5x10^-4, K_a(CH3COOH) = 1.8x10^-5, K_b(NH3) = 1.8x10^-5. JEE Main level
Equal volumes of 0.20 M NH3 and 0.20 M HCN are mixed at 298 K. K_b(NH3) = 1.8x10^-5, K_a(HCN) = 4.0x10^-10. Treat the resulting salt solution with pH = 7 + tfra... JEE Advanced level
30.0 mL of 0.20 M HCOOH is mixed with 30.0 mL of 0.20 M NaOH at 298 K. K_a(HCOOH) = 1.8x10^-4. Use pH = 7 + tfrac12(pK_a + log c). The pH of the mixture is JEE Advanced level
0.10 M solutions of the sodium salts NaA and NaB of two weak acids HA and HB have pH 9.00 and 10.00 respectively at 298 K. The ratio K_a(HA)/K_a(HB) is JEE Advanced level
A 0.050 M solution of the chloride BCl of a weak base BOH has pH 5.45 at 298 K. Using K_h = [H+]^2/c (valid for small hydrolysis), find K_b of BOH and report it... JEE Advanced level
Sodium phenoxide, C6H5ONa, is 1.0x10^-3 M in water at 298 K. For phenol K_a = 1.0x10^-10. Because hydrolysis is not small, solve K_h = x^2/(c - x) exactly as a ... JEE Advanced level
For a 0.10 M solution of sodium acetate at 298 K (K_a = 1.8x10^-5, h = sqrtK_h/c), the solution is diluted tenfold with pure water. Select the correct statement... JEE Advanced level
In which of the following cases is the final solution (298 K) basic, i.e. pH > 7? Data: K_a(CH3COOH) = 1.8x10^-5, K_b(NH3) = 1.8x10^-5, K_a(HCN) = 4.0x10^-10. JEE Advanced level
What is the pH of a 0.10 M aqueous solution of CH3COONa? (K_a of acetic acid = 1.8x 10^-5; use pH = 7 + tfrac12(pK_a + log C)) NEET level
The aqueous solution of which of the following salts has pH less than 7 at 298 K? NEET level
The base dissociation constant of ammonia is K_b = 1.8x 10^-5 and K_w = 1.0x 10^-14. What is the hydrolysis constant K_h of NH4+ ? NEET level
For 0.010 M NaCN (K_a of HCN = 4.9x 10^-10), the percentage degree of hydrolysis, using h = sqrtK_h/C, is closest to NEET level
Calculate the pH of 0.20 M NH4Cl at 298 K. (K_b of NH3 = 1.8x 10^-5; use pH = 7 - tfrac12(pK_b + log C)) JEE Main level
0.05 M NH4CN is prepared. Given pK_a(HCN) = 9.3 and pK_b(NH3) = 4.76, and using pH = 7 + tfrac12(pK_a - pK_b) for the salt of a weak acid and a weak base, the p... JEE Main level
The pH of 0.10 M CH3COONa (K_a = 1.8x 10^-5) is calculated by the usual hydrolysis formula. The solution is now diluted to 100 times its volume with water. Assu... JEE Main level
What is [OH-] in 0.20 M HCOONa? (K_a of formic acid = 1.8x 10^-4) JEE Main level
Find the pH of 0.10 M NaF at 298 K. (K_a of HF = 6.8x 10^-4; use pH = 7 + tfrac12(pK_a + log C).) Give the answer to two decimals. JEE Main level
The degree of hydrolysis h of 0.020 M NH4Cl (K_b = 1.8x 10^-5) is calculated from h = sqrtK_h/C. Report hx 10^4 to two decimals. JEE Main level
Consider 0.10 M solutions of three salts: NaA, BCl and BA, where HA is a weak acid with K_a = 1.0x 10^-5 and BOH is a weak base with K_b = 1.0x 10^-5. Use the s... JEE Main level
100 mL of 0.20 M NH4Cl is mixed with 100 mL of 0.20 M NaCN. K_a(HCN) = 4.9x 10^-10, K_b(NH3) = 1.8x 10^-5. Assume that the equilibrium NH4+ + CN- <=> NH3 + HCN ... JEE Advanced level
100 mL of 0.10 M NaHCO3 is mixed with 100 mL of 0.10 M NaOH. For carbonic acid pK_a1 = 6.35 and pK_a2 = 10.33. Neglect the second hydrolysis step of the carbona... JEE Advanced level
For carbonate hydrolysis, CO3^2- + H2O <=> HCO3- + OH-, K_h = K_w/K_a2 with K_a2(H2CO3) = 4.7x 10^-11. Neglect the second step. Find the percentage degree of hy... JEE Advanced level
Ammonium formate, HCOONH4, is a salt of a weak acid (K_a = 1.8x 10^-4) and a weak base (K_b = 1.8x 10^-5). Use h = dfracsqrtK_h1+sqrtK_h with K_h = K_w/(K_a K_b... JEE Advanced level
Three sodium salts NaA, NaB and NaC are made up as 0.10 M solutions. The conjugate acids have K_a values 1.0x 10^-4 (HA), 1.0x 10^-6 (HB) and 1.0x 10^-8 (HC). U... JEE Advanced level
A 0.20 M solution of NaA (HA: K_a = 2.0x 10^-5) is diluted to 0.0020 M at 298 K. Use h = sqrtK_h/C and [OH-] = sqrtK_h C. Which statements are correct? JEE Advanced level
100 mL of 0.20 M CH3COONa (pK_a = 4.74 for acetic acid) has pH = 9.02. Which of the following changes give the stated pH? (Assume additive volumes; for the exce... JEE Advanced level