A 29.9 mL sample of 0.342 M ammonia, NH3, is titrated with 0.387 M nitric acid. At the equivalence point, the pH is . Use the Tables link in the References for any equilibrium constants that are required. Submit AnswerRetry Entire Group9 more group attempts remaining

At the equivalence point, the pH is 5

Explanation:

Step 1: Data given

Volume of 0.342 M ammonia NH3 solution = 29.9 mL = 0.0299 L

Molarity of the nitric acid (HNO3) solution = 0.387 M

Step 2: The balanced equation

NH3 + HNO3 → NH4+ + NO3-

Step 2: Calculate moles NH3

Moles NH3 = molarity * volume

Moles NH3 = 0.342 M * 0.0299 L

Moles NH3 = 0.010226 moles

Step 3: Calculate moles HNO3

For 1 mol NH3 we need 1 mol HNO3 to produce 1 mol NH4+ and 1 mol NO3-

For 0.010226 moles NH3 we need 0.010226 moles HNO3

Step 4: Calculate volume HNO3

Volume = moles / molarity

Volume = HNO3 = 0.010226 moles / 0.387 M

Volume = 0.0264 L = 26.4 mL

Step 5: Calculate total volume

Total volume = 29.9 mL + 26.4 mL = 56.3 mL

Step 6: Calculate molarity NH4+

Both NH3 and HNO3 will be consumed. There will be produced 0.010226 moles NH4+

[NH4+]= 0.010226 moles / 0.0563 L

[NH4+] = 0.182 M

Step 7: The equation

NH4+ ⇆ H+ + NH3

Step 7:  Define Ka

Ka = [H+] *[NH3] / [NH4+]

Ka = X² / 0.182 M

5.6 * 10^-10 = X² / 0.182 M

X²  = 1.02 *10^-10

X = 1*10^-5 M = [H+]

Step8: Calculate pH

pH = -log[H+]

pH = -log[10^-5]

pH = 5

At the equivalence point, the pH is 5