When ammonium (NH4+) is placed in water, a chemical reaction occurs and an equilibrium is established as shown below.   NH4+ (aq) + H2O (l)  ⇄  NH3 (aq) + H3O+ (aq) The pKa of NH4+ is 9.25.  Predict whether the acid form (NH4+) or the base form (NH3) is predominant at pH = 7.4.

Identify the reactant that is a Brønsted-Lowry acid and the reactant that is a Brønsted-Lowry base in each of the following: Part C 2- CO3² (aq) + H₂O = HCO3(aq) + OH-(aq) H₂O is the acid (proton donor); CO3² is the base (proton acceptor). H₂O is the acid (proton acceptor); CO² is the base (proton donor). O CO3²- is the acid (proton acceptor); H₂O is the base (proton donor). O CO² is the acid (proton donor); H₂O is the base (proton acceptor). Submit Part B Request Answer H₂SO4 (aq) + H₂O(1)→ H3O+ (aq) + HSO4 (aq) O H₂O is the acid (proton donor); H₂SO4 is the base (proton acceptor). O H₂SO4 is the acid (proton acceptor); H₂O is the base (proton donor). O H₂SO4 is the acid (proton donor); H₂O is the base (proton acceptor). O H₂O is the acid (proton acceptor); H₂SO4 is the base (proton donor). CH3-COO (aq) + H3O+ (aq) = H₂O(1) + CH3-COOH(aq) CH3-COO is the acid (proton acceptor); H3O+ is the base (proton donor). O H3O+ is the acid (proton donor); CH3-COO is the base (proton acceptor). O…

Phosphoric acid, H3PO4, is a triprotic acid with Ka1 = 7.5 x 10-3, Ka2 = 6.3 x 10-8 and Ka3 = 4.8 x 10-13. What is the equilibrium constant for the reaction:      HPO42-(aq) + H2O (ℓ) ⇆ H3O+ (aq)+ PO43-(aq)

[Review Topics] [References] Use the References to access important values if needed for this question. Identify whether each species functions as a Brønsted- Lowry acid or a Brønsted-Lowry base in this net ionic equation. F™ Brønsted- Lowry A + H₂CO3 (aq) = Brønsted- Lowry In this reaction: The formula for the conjugate The formula for the conjugate H₂CO3 is HF (aq) + HCO3 Brønsted- Brønsted- Submit Answer 9 more group attempts remaining Lowry Retry Entire Group of Fis of Lowry 00 F12

Calculate the pH of a 0.123 M solution of ethylenediamine (H₂NCH₂CH₂NH₂). The pKą values for the acidic form of ethylenediamine (H3NCH₂CH₂NH³) are 6.848 (pKal) and 9.928 (pKa2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H₂NCH₂CH₂NH₂] : = [H₂NCH₂CH₂NH3] = M M

Hydrogen phosphate ion is an anion with the chemical formula H2PO4–. In the aqueous state, it undergoes the following reactions: (i) H2PO4– (aq) + H2O (ℓ) ⇌ H3O+ (aq) + HPO42– (aq) (ii) H2PO4– (aq) + H3O+ (aq) ⇌ H3PO4 (aq) + H2O (ℓ) Which of the following statements is true about the above reactions?   Question 2 options:   In reaction (i) hydrogen phosphate acts as a Brønsted-Lowry base and in reaction (ii) it acts as a Brønsted-Lowry acid.   In reaction (i) and (ii) hydrogen phosphate acts as a Brønsted-Lowry base.   In reaction (i) hydrogen phosphate acts as a Brønsted-Lowry acid and in reaction (ii) it acts as a Brønsted-Lowry base.   In reaction (i) and (ii) hydrogen phosphate acts as a Brønsted-Lowry acid.

Calculate the pH of the solution after the addition of each of the given amounts of 0.0651 M HNO3 to a 50.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04. What is pH after adding 62.6 mL HNO3?

Some flowering plants will change color based on the pH of the soil.  For example, hydrangeas will flower in blue/purple shades under more acidic conditions and in pink/red shades under more basic conditions.  Consider NaHPO3 and NaHPO4.  Which compound, when added to soil, would be the better choice to cause hydrangeas to favor blue/purple shades?  Explain your reasoning.

The pKa of acetic acid is 4.75. What is the ratio of the concentration of sodium acetate to the concentration of acetic acid at pH 5.75? equation: ch3cooh (aq)+oh-(aq) = ch3coo-(aq)+h20(l)

The following equilibrium constants have been dtermined for oxalic acid at 25°C: H2C2O4 (aq) = H+ (aq) + HC2O4- (aq)                           K1= 6.5 x 10-2 H+ (aq) + C2O42- (aq) = HC2O4- (aq)                            K2= 1.6 x 104 Calculate the equilibrium constant for the following reaction at the same temperature: H2C2O4 (aq) = 2H+ (aq) + C2O42- (aq)                          Kc= ?