Question

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 27.0 mL of HNO3. Express your answer numerically.

A 52.0-mL volume of 0.35 MM CH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 15.0 mL of NaOH.

and

Imagine that you are in chemistry lab and need to make 1.00 LL of a solution with a pH of 2.80.

You have in front of you

• 100 mL of 7.00×10⁻² MM HCl,
• 100 mL of 5.00×10⁻² MM NaOH, and
• plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 84.0 mL of HCl and 88.0 mL of NaOH left in their original containers

Assuming the final solution will be diluted to 1.00 LL , how much more HCl should you add to achieve the desired pH?

Answer 1

Ž NHN Answer: Nu + nno Loctially 0.015 moles 0.0135 After ok? 0.015 – 0.0135 o . 0.0135 after &k" = 0.015-0.0135 Conch of Nu 102 = 1.47x10-S cone of NH No after ? = 0.0135 M = 1.32*10 102 pon - plezt log [ 1.32x10"] [1.47x105] - 4.74 +0.95 pon = 5.69 put pon = 14 {pn = 8.31] chi Jonitially After rich choon + Naon chooNa + no 0.0182 moles 0.006 0.0182 -0.006 0.006 conc" of cncoon after 6 = 1.820x10 4 " " choona " " = 8.95x105 ри = rea • La9, Гchoors. ) - 4.74 + (-0.308) Jpn = 4:43 pn = peas (encoon]

(011) nel + Noon reteally 1.12 moles 0-6 moles 10-3 Change 1.12 + vxo.o7 – 0.6 0 Nad the o 0.6 xio C after addition and ok = 1.12 +0.07xV - 0-6 1.52 +0.07x v moles of hcl x 103 conch of hd = 0.524 0.0pxv M ph= -log[ut 2.8 = -69, [u'] [nt] = 1.58x10² M 80 X10 3 0.52 +0.07xv = 1.58x103 0.07*V = 1.06x103 [v=15.14 m2 The volume of siel that should be added is 15.14 m)