Question

You create solutions of H₂SO₄ and NaOH with concentrations of 1.23 and 0.81,respectively. If you titrate 10.0 mL of the H₂SO₄ solution with the NaOH base you have created, at what volume do you expect to see the equivalence point?
mL NaOH =

If the actual mL used is 29.6, what was the actual concentration of the base assuming the acid concentration was correct? Actual [NaOH] =____M

If the actual mL used is 29.6, what was the actual concentration of the acid assuming the base concentration was correct? Actual [H₂SO₄ ] =_____M

What is the %error (+ or -) in each case?
%Err = (Actual - Given)/Actual x 100%
%error [NaOH] =

%error [H₂SO₄ ] =

Answer 1

Solution i 1st part H₂SO4 NaOH 2 + sor Nat + oH – H₂SO4 + 2NaOH - Na, soy + 2H20 H₂Soy H₂soy is a diabasic alid. So, 1 mole of will neutralize a mole of Naot solution. given date, concentration of H₂SO4, M = 1123 M. 1.23 M H₂SO4 = 2x1.23N H₂SO4 - 2.46 NH₂SO4. volume À H₂SO4 = 10 ml Concentration st NapH, M₂ = 0.81 M. 081M Nash - 0.81 N NaOH. het, me vole of HapH = V2 We have, NV - NqV2 (H₂SO4) (PaoH) 2.464 10 = 0.81 XV₂ -

2.46410 0.81 - 30.37 ml . volume of Naot – 30.37 ml. 2nd part & actual volume of NaOH is, - 29.6 ml. concentration of H₂Soy is correct ;, e. M = 1.23M and the . N = 2X1023 N. - 2.46 N. We have, NV = N₂ V₂ (H2SO4) (NaOH) N = N. V 2.46410 29.6 = 0'83 N. N NOW, 0.83 N NeoH = 0.83 M N2OH. Naoth-0.83M. i actual concentration of 3rd part 3rd Paul and actual volm of HapH is, &- 29.6 me. concentration of NaOH (M2) = 0.81M is correct. N₂ = 0.81 0.

We have N.V. - N2 V2 (H₂SO4) (Nash) N = N2 X V₂ 0.818 29.6 N = 2.4 N. 2.44 H₂ Soy - 24 M H₂SO4 for dibasić . seid M = 28. - 1.2M H₂SO4 N- M Actual concentration of H₂ soy is 1.2M. N=ntactory M. here, n=2 for H₂SO4. here denote actual - given basicity of H₂SO4 % error (NaOH) = armud X100% 100% 0.83 - 0.81 0.83 - X - 2.41% Actual - given yercoy / H, S X 100% Actual = 12-1:23 X 100 1.2 – 2.5 1. ms