Question

A solution of potassium hydroxide

(KOH) was titrated against a solution
of hydrochloric acid. It took 35 cm³
of the hydrochloric acid to
completely neutralise 50 cm³ of
potassium hydroxide. Work out the
concentration of the potassium
hydroxide solution in mol/dm³ if the
concentration of the acid was 2
mol/dm³. Round your answer to 1
d.p.

Answer 1

first you'd need to write the balanced symbol equation of the reaction between KOH and HCL which would be:

KOH+HCL-> KCL+H20

you know that 35cm^3 of 2 mol/dm^3 HCL has been used, and you need to work out the moles of the HCL from this info using the equation:

concentration(mol/dm^3 )=moles(mol)/volume(dm^3)

rearranging the equation to make moles the subject, you'd get: moles=concentration x volume

plugging in the info, you'd get moles of HCL= 35/1000 dm^3 x 2 moldm^3= 0.07 mol

because there's a 1:1 ratio of HCL:KOH, the moles of KOH would also equal 0.07 mol

now using the con equation, you'd get the concentration of KOH= 0.07mol/(50/1000) dm^3= 1.4 moldm^3

Step-by-step explanation:

hope this helps and sorry if any of my calculations are wrong! you may want to double check all the calculations.