A solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh.

a solution of hydrochloric acid of unknown concentration was titrated with 0.10M naoh. if a 100.ml sample of the hcl solution required exactly 10.ml of the naoh solution to reach the equivalence point, what was the ph of the hcl solution?

M1V1 = M2V2

(0.10 mol/L) (10 mL) = (x) (100 mL)

x = 0.010 M

pH = log [H+] = - log 0.01 = 2

"its obviously not negative one because ph isnt negative."

Wat's the pH of 2.00 M HC?

pH = -log 2.00 = -0.301

pH values can be negative. They are not often used, but they can exist.

Actually very concentrated strong acids can achieve a negative ph, consider the pH of 10M HCl:

-log(10) = -1

Anyway, back to the question:

You know how many moles of NaOH were titrated, and since the reaction:

H+ + OH- ---> HOH (H2O if you prefer)

has 1:1 stochiometry, you can use the conversion of # moles OH- titrated = # moles H+ consumed and vice versa.

.1M * .01L = .001 moles NaOH used.

.001 moles OH- = .001 moles H+, so:

.001/.1L = .01M

-log(.01) = 2

The answer is thus b.

0.10M of NaOH
Multiply the molarity of NaOH by the volume of NaOH
Molarity is measured in litres, volume has to be in litres too, hence I divided the mL by 1000 to get L (1L = 1000mL)
0.10M x 10 / 1000 = 0.001,mols of NaOH
Ratio 1:1
0.001mols of HCl
Divide the mols of HCl by the volume of HCl
0.001 / 100 / 1000 = 0.01M

Use the -log() formula
pH = -log(0.01)
pH = 2
Your answer is B