Given 20 mL NaOH; 0.1 M, add 0.05 M HCl, 10 mL at a time.

What is the pH initially of NaOH?
What is the pH after adding 10 mL 0.05 M HCl?
What is the pH after adding 20 mL 0.05 M HCl?
What is the pH after adding 30 mL 0.05 M HCl?
What is the pH after adding 40 mL 0.05 M HCl?
What is the pH after adding 50 mL 0.05 M HCl?
What is the pH at equivalence point?

This is the equivalent of an afternoon's work. What do you not understand about this and I can help you through it. The secret to this problem is to recognize what you have in the solution. For example, to start, you have a solution of NaOH, 0.1 M so the pOH is 1 and pH = 13.

a (s) is obtained by mixing 25ml of 0.5 mol/l HCl and 25ml of 0.5 mol/l H2SO4 solution is titrated with 93.75ml of barium hydroxide solution

1-write the ionic eq.
2- calculate C
3. determine mass of barium sulfate obtained:P
*molar masses: Ba=137
S=32 O=16

To answer these questions, we need to calculate the concentrations of NaOH and HCl after each addition and then determine the pH using the concept of neutralization.

First, let's find the initial concentration of NaOH using its molarity and volume:

Initial moles of NaOH = molarity * volume
= 0.1 M * 20 mL
= 2 mmol

Initial concentration of NaOH = moles/volume
= 2 mmol/20 mL
= 0.1 M

Now, let's calculate the pH initially using the concentration of NaOH:

pH = -log[H+]

In this case, since NaOH is a strong base, it fully dissociates into Na+ and OH- ions. Therefore, the concentration of [OH-] is equal to the concentration of NaOH.

pOH (initially) = -log[OH-]
= -log(0.1 M)
= 1

pH (initially) = 14 - pOH
= 14 - 1
= 13

So, the initial pH of the NaOH solution is 13.

Moving forward, let's proceed with calculating the pH after adding the HCl solution, 10 mL at a time.

In the first step, we add 10 mL of 0.05 M HCl to the 20 mL of 0.1 M NaOH:

Moles of HCl added = molarity * volume
= 0.05 M * 10 mL
= 0.5 mmol

Moles of NaOH remaining = Initial moles of NaOH - Moles of HCl added
= 2 mmol - 0.5 mmol
= 1.5 mmol

Volume of the final solution = Initial volume of NaOH + Volume of HCl added
= 20 mL + 10 mL
= 30 mL

Concentration of the final solution = Moles of NaOH remaining / Volume of the final solution
= 1.5 mmol / 30 mL
= 0.05 M

Now, let's calculate the pH after adding 10 mL 0.05 M HCl:

Moles of OH- remaining = Moles of NaOH remaining
= 1.5 mmol

Concentration of OH- remaining = Moles of OH- remaining / Volume of the final solution
= 1.5 mmol / 30 mL
= 0.05 M

pOH (after adding 10 mL HCl) = -log[OH- remaining]
= -log(0.05 M)
= 1.3

pH (after adding 10 mL HCl) = 14 - pOH (after adding 10 mL HCl)
= 14 - 1.3
= 12.7

So, the pH after adding 10 mL of 0.05 M HCl is 12.7.

Now, you can repeat the same steps for the remaining additions of HCl (20 mL, 30 mL, 40 mL, and 50 mL) to find the pH after each addition.

Finally, the pH at the equivalence point can be determined by calculating the remaining moles of NaOH and finding the concentration and pH using the same steps as before. At the equivalence point, the moles of acid added (HCl) will be equal to the moles of base present (NaOH), resulting in a neutral solution with a pH of 7.