The pH of an unknown solution is 1.04. Calculate the [H,o for this solution. (M)

Question

The pH of an unknown solution is 1.04. Calculate the [H,o for this solution. (M) Question 25 A 0.0250 M sample of vinegar, which is an aqueous solution of acetic acid, CH3COOH, requires 16.2 mL of 0.500 M NaOH to reach the endpoint in a titration. What volume (mL) of the acetic acid was in the original solution? CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) Question 26 Calculate the [H3O+] of each aqueous solution with the following [OH- NaOH, 1.0 x 102 M aspirin, 1.8 x 10-11 M A. 4.0e-9 M B. 1.0e-9 M C. 5.6e-4M D. 1.0e-12 M E/milk of magnesia, 1.0 x 10-5 M sea water, 2.5 x 10-6 M

Explanation / Answer

1.concentration of H+ = concentration of H3O+

ph = -log(concentration)

1.04= ph= -log (H+)

log(H+)=-1.04

H+ = 9.1 *10^-2

2. for the above equation 1 mole acid is treated by 1 mole of base

so ,

so molarity * volume of NaOH = molarity * volume of CH3COOH

0.5 * 16.2 = 0.0250 * vol

volume = 8.1/0.0250 = 324 ml .

3. Kw = Kacid * Kbase

Kw =10^-14(for NaOH = 1.0 * 10^-2) means OH = 10^-2

10^-14/1.0* 10^-2= 1.0 * 10^ -12 M.......

for aspirin

10^-14 / 1.8 * 10^-11 = 5.6 *10^-4M

for milk of magnesia

10^-14/1*10^-5 = 1.0 * 10^-9 M

for sea water

10^-14 / 2.5 *10^-6 = 4*10^-9M

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