what volume of 0.25M NaOH is required to neutralize 500mL of HCl solution with pH 0.83?

Question

households acids/bases experiment

Answer 1

Convert pH to concentration of H+ ions (or molarity) then use the equation:
M1 x V1 = M2 x V2

Answer 2

Well, if you have pH 0.83 acid in your household, I hope it is in a safe place. You solve this problem with the equation
( V * M ) acid = (V * M ) base.
We want to find Vbase. We can find Macid from the pH, which is what we do first.
[H+] = 10-pH = 10^-1 * 10^0.17 This is appx 0.15. Since HCl totally dissocates in water, 0.15 is Macid. Now to the equation:
( 500 x 0.15) = (V * 0.25), Vbase = 300 mL.

Answer 3

First you need to know the concentration of the acid. As all that is given is the pH you need to work it out from there.
pH = -log[H+]
So [H+] = 10^(-pH)
= 10^(-0.83)
=0.15
Now convert the conc. to moles
mole H+ = vol (in litres) * [H+]
= 0.5 * 0.15
= 0.075

Balanced equation is
NaOH + HCl = NaCl + H2O
As it is a 1:1 relationship between the NaOH and HCl you need 0.075 moles of NaOH to neutralise the HCl.

Vol NaOH = moles NaOH / [NaOH]
= 0.075/0.25
= 0.30 l
= 300ml

NOTE: [ ] refers to concentration

Answer 4

HCl pH = 0.83 = 83. 10^-2
-2

[H]+ = 2 - log 83 M
moles of HCl = (2-log 83 M) x 500mL = @ mmoles

NaOH + HCl --> NaCl + H2O

so, volume of NaOH must be:
@ / 0.25M = ....mL

Answer 5

pH=-log[ H+ molar concentration/ liter];which in this case=0.83
so the H+ concentration is 10^-0.83= 0.148 moles H+/liter;
since were dealing with 1/2 liter or 500 ml. we have 0.078 moles
of H+ ion. Since a 0.25 M NaOH contains 0.25 moles OH-/liter
or 0.00025 moles OH-/ ml. The answer is .078/0.00025 = 312 mls of a
0.25M NaOH to bring a 500 ml solution of HCL pH0.83 to a
neutral pH=7.0

Answer 6

pH = 0.83 means -log[H+] = 0.83, or [H+] = 10^-0.83 (here [H+] means molar concentration of hydrogen ions). In one liter there are that many ions, so in half a liter, the figure is 10^-0.83/2.
We need that many hydroxyl ions to neutralize the solution. We have 0.25 M NaOH, which means there are 0.25 moles of NaOH in one liter. So there will be 10^-0.83/2 moles of OH- in 4x10-0.83/2 liters.