Question

In: Chemistry

A.) What is the pH of a buffer prepared by adding 0.607 mol of the weak...

A.) What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA to 0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. ANSWER IS: 6.072

B.) What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. ANSWER IS: 5.776

C.) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

I had to request the answer for part B but I don't know how to work the problem out. Please help me!

Expert Solution

A.) What is the pH of a buffer prepared by adding 0.607 mol of the weak acid HA to 0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7. ANSWER IS: 6.072

Solution :- Weak acid and its conjugate base moles are given and asked to calculate the pH

Lets first calculate the pH of the solution using the Henderson equation

pH= pka + log ([base]/[acid])

pka = -log ka

= -log 5.66*10^-7

= 6.247

Now lets put the values in the formula

pH= 6.247 + log [0.406/0.607]

pH= 6.247+(-0.175)

pH= 6.072

B.) What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid. ANSWER IS: 5.776

Solution :-

After adding the 0.150 mol of HCl the acid moles will increase and base moles will decrease by the same amount

Therefore new moles of acid = 0.607 +0.150 = 0.757 mol

New moles of base = 0.406-0.150 = 0.256 mol

Now lets calculate the pH using the same formula

pH= 6.247 + log [0.256/0.757]

pH= 6.247+(-0.471)

pH= 5.78

C.) What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Solution :-

After adding the NaOH the moles of acid will decrease and moles of base will increase

So new moles of acid = 0.607 – 0.195 = 0.412 mol

Moles of base = 0.406 +0.195 = 0.601 mol

Now lets calculate the pH

pH= 6.247 + log [0.601/0.412]

pH= 6.247+ 0.164

pH= 6.41

queen_honey_blossom answered 1 month ago

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