Question
A weak acid HA (pka=6.00) was titrated with 1.00M NaOh. The acid solution had a volume of 100ml and a concentration of .100M. Find the pH at the following volumes of added base. Vb=20ml
Answers
millimols acid = mL x M = 100 x 0.1 = 10
millimols NaOH = 20 x 1 = 20
So all of the acid is neutralized and y ou have 10 mmols NaOH in excess.
So (OH^-) = mmols/total mL = 10/200 = 0.1
pOH must be 1 so pH must be 13.
millimols NaOH = 20 x 1 = 20
So all of the acid is neutralized and y ou have 10 mmols NaOH in excess.
So (OH^-) = mmols/total mL = 10/200 = 0.1
pOH must be 1 so pH must be 13.
Note that 10/200 is not 0.1 but is ?
Then pOH = -log(OH^-). Calculate pOH.
Then pH + pOH = pKw = 14.
You know pKw and you know pOH, solve for pH.
Then pOH = -log(OH^-). Calculate pOH.
Then pH + pOH = pKw = 14.
You know pKw and you know pOH, solve for pH.
pH=5.41 after adding 20ml of 0.100M NaOH into 100ml of 0.100M HA
I would have posted my solution, but the site monitor keeps kicking out my post as a website post.
This is an acid/base titration with a common ion. Hope you can get a solution.
This is an acid/base titration with a common ion. Hope you can get a solution.
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