Question

What would the pH be of a solution obtained by mixing 0.2 mL 1 M HCl with 10 mL distilled water? What would the pH be of a solution obtained by mixing 0.2 mL 1 M HCl with 10 nM NaOH (aq)? How do these values compare to the pH values you obtained when you added acid to your buffers?

Answer 1

V = 0.2 ml of M = 1 M of Hcl

M1V1 = M2V2

M2 = M1V1/V2 = 1*(0.2)/(10+0.2) = 0.01960784313 M o fHcl

[H+] = 0.01960784313 M

pH = -log(H+) = -log(0.01960784313) = 1.7075

b)

NOTE: assume "nM" is not nanoMolarity but mL, nano is not used in lab; is extremely small

pretty similar to HCl:

V = 0.2 ml of M = 1 M of NAOH

M1V1 = M2V2

M2 = M1V1/V2 = 1*(0.2)/(10+0.2) = 0.01960784313 M o NAOH

[OH-] = 0.01960784313 M

pOH = -log(OH-) = -log(0.01960784313) = 1.7075

pH = 14-1.7075 =12.2925

c)

When you add acid/base to a buffer, they will not change drastically i.e. pH is pretty near to the pKa vlaues