find the Ph of the mixture when 25 cm^3 of 2 moldm^-3 sodium hydroxide solution a added to 50cm^3 of 2 moldm^-3 ethanoic acid. Ka =1.7x 10^-5. I thought the answer was 2.5 but apparently it's 4.8 could you help explain this please

Question

jfraser · Answer

You need to write a blanced reaction between the $NaOH$ and the $HC_2H_3O_2$ first

Then you have a limiting reactant problem, where you have different amounts of the two reactants.

The $leftover$ reactant will then use the $K_A$ equation to find the pH

jami1e123 · Answer

That's what i did. As the concentrations are the same i worked out that there will be 25cm^3 of 2 moldm^-3 ethanoic acid remaining which means that there are 1/20 mol of ethanoic acid leftover in a solution on 75cm^3 which makes the concentration 2/3 moldm^-3. using the Ka equation this gave me 2.5 as -log( the square root of 1.7x10^-5 x2/3) =2.5

jfraser · Answer

without doing it out myself, i'd guess that your concentration to mole conversions are off, from step 1.

I find it difficult to compare $concentrations$, so we usually convert them to $moles$ first, then do the limiting reactant stoichiometry

jfraser · Answer

so 0.025L of 2M $NaOH$ is .005mol of $NaOH$

and 0.050L of 2M $HC_2H_3O_2$ is 0.010mol of $HC_2H_3O_2$ so you'll have 0.005mol of ethanoic acid left over

jfraser · Answer

0.005 isn't 1/20, it's 1/200

jami1e123 · Answer

I don't think thats correct concentration= mol/vol and 0.005/0.025= 0.2 not 2

jfraser · Answer

$$ 0.050 \cancel{L \space HC_2H_3O_2} * \frac{2mol \space HC_2H_3O_2}{\cancel{1L \space HC_2H_3O_2}} = 0.100mol \space HC_2H_3O_2$$which is the amount of ethanoic acid you have at the start. 

you'll use half of it, so you'll be left with 0.050 moles of ethanoic acid once all the sodium hydroxide has been used

jfraser · Answer

You're right, I added too many zeros

jfraser · Answer

that 0.050 moles of acid is now in 75mL of solution, because they've been added, so the new "starting" concentration of the acid for its equilibrium expression is 
$$\frac{0.050mol \space HC_2H_3O_2}{0.075L \space HC_2H_3O_2} = 0.667M \space HC_2H_3O_2$$

jami1e123 · Answer

so 1.7x10^-5= H+^2 / 0.667 so H+= 3.3665x10^-3 and -log3.3665x10^-3
= 2.5 ?

jfraser · Answer

that's what I get, unfortunately. I don't know what to tell you

jfraser · Answer

If you forget to square root the $[H^{+1}]$, you'll get a "pH" of about 4.9, which may explain it

jfraser · Answer

I'd guess someone just made a mistake on their answer key. 2.5 is the correct pH

jami1e123 · Answer

alright thank you