How many milliliters of 0.250 M KoH will be needed to titrate 12.53 mL of 0.130 M #HNO_3#?

2 Answers
Apr 17, 2017

Answer:

You will need 6.516 mL of KOH to reach the equivalence point.

Explanation:

Since KOH has only one #OH^-# ion in its formula, and #HNO_3# is monoprotic (donates only one hydrogen ion), the titration reaches an equivalence point when

Moles #HNO_3# = moles #KOH#

And since moles (of solute) = concentration x volume, we can write

#M_a*V_a=M_b*V_b#

(the #a# and #b# representing the acid and base, respectively)

We know the first three quantities in this equation. We find the fourth (#V_b#) as follows:

#(0.130)*(0.01253)=(0.250)V_b#

#V_b =((0.130)*(0.01253))/(0.250) = 0.006516 L#

or #6.516 mL#

Apr 17, 2017

Answer:

Approx. #6-7*mL..........#

Explanation:

We need (i) a stoichiometric equation........

#HNO_3(aq) + KOH(aq) rarr KNO_3(aq) + H_2O(l)#

And then (ii) equivalent quantities of potassium hydroxide and nitric acid:

#"Moles of nitric acid"=12.53*mLxx10^-3L*mL^-1xx0.130*mol*L^-1=1.63xx10^-3mol#.

And thus we need #(1.63xx10^-3mol)/(0.250*mol*L^-1)xx10^3*mL*L^-1=6.52*mL#