When the reversible reaction HC;H;0, (aq) > H'(aq) + C;H;O2(aq) is at equilibrium at room temperature, pH of the reaction mixture is 5. What will be the change in pH when you add a large amount of NaOH (aq) to the reaction mixture? a) pH will not change because NaOH is not a part of the reaction equation. b) pH will not change because acetic acid and NaOH form a buffer. c) pH will increase because NaOH will completely neutralize acetic acid. d) pH will decrease because NaOH will completely neutralize acetic acid. In the reversible reaction A (aq) + B(aq) + → C(aq), reactant A is very expensive. What are two ways to get it to react as fully as possible to form as much C as possible? a) remove C as it forms, and use an excess of A b) add another reactant that will form a precipitate with B c) remove C as it forms, and use an excess of B d) add another reactant that will form a precipitate with A

Consider the following equilibrium. Would it lie more towards the reactants or the products? Why?  HCO3-(aq) + SO42-(aq) --> HSO4-(aq) + CO3-2(aq)

answer and explain plz

Question 12 The equilibrium for the reaction between (CH,).,NH, a weak base, and water is represented by the equation below. The table shows the pH of three solutions of (CH,),NH(aq) at 25° C. (CH3),NH(aq) + H,0(1) 3D (CH,),NH;" (aq) + ОH (ад) К, — 5.4 х 10 -4 at 25°C [(CH3),NH] pH at 25°C 0.050 11.69 0.10 11.85 0.20 12.01 Based on the information given, which of the following is true? Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach A 21 + equilibrium a smaller amount of the conjugate acid (CH,),NH2™ is produced. Solutions with a higher concentration of (CH,),NH have a higher pOH because to reach equilibrium more OH¯ is produced. 3/21 Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium a smaller amount of the conjugate acid (CH,),NH2™ is produced. Solutions with a higher concentration of (CH,),NH have a higher pH because to reach equilibrium more OH is produced. 3/2-

The compound methylamine is a weak base like ammonia. A solution contains 5.89x10-2 M CH3NH3+ and 0.330 M methylamine, CH3NH2. The pH of this solution is

Please help:   A)  Write the acidic equilibrium equation for HC₆H₅CO₂ B)  The concentration of hydroxide ion in an aqueous solution at 25°C is 5.7 × 10⁻⁴ M. What is the concentration of the hydronium ion? C)  The concentration of the hydronium ion in an aqueous solution at 25 °C is 0.0025 M. What is the concentration of the hydroxide ion? D)  The OH⁻ concentration in an aqueous solution at 25 °C is 8.3 × 10⁻³. What is [H⁺]?   Thank you!!

3c.) Suppose it is desired to maintain the pH of a solution at 4.2 using only acetic acid and sodium acetate. Would the concentration of acetic acid or sodium acetate have to be higher in the solution? O sodium acetate O acetic acid

(a) Using the expression Ka=[H+][A−]/[HA], explain how to determine which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq). Do not perform any numerical calculations. (b) Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq) ? Justify your answer including the calculation of percent ionization for each solution.

Drinking water is often disinfected with Cl₂, which hy-drolyzes to form HClO, a weak acid but powerful disinfectant Cl₂(aq)+2H₂O(l) ⇌HClO(aq)+H₃O(aq)+Cl(aq)The fraction of HClO in solution is defined (a) What is the fraction of HClO at pH 7.00 (Kaof HClO = 2.9X10⁻⁸)? (b) What is the fraction at pH 10.00?

HA is a weak acid. Select the correct dropdown value to correctly complete the incomplete concentration table below being used to determine the equilibrium concentrations of HA, H3O*, and A¯ of a 0.1300 M HA solution that has a pH of 3.60 at 25°C. HA(aq) + H,0(1) = H,0^(aq) + A-(aq) Initial 0.1300 Change Equilibrium (b) (a] The value for a is [ Select ] The value for b is [ Select ] What is the Ka of this weak acid? [ Select]