Question

QUESTIONS 1. A student masse 0878 of an unknown acid follows the titration procedure required to reach the endpoint on a 10.0 ml Calculate the molar mass of the acid. masses out 0.878 of die in 100 ml volume and titration procedure of Partsherds th the average volume of Non to reach the endpoint on 100 ml aliquot of the acid solution is 18.3 m . 2. In the vinegar titrations (Part A), you used 1.00 mL sample of the vinegar. How would the required volume of titrant change increase/decrease/no change) under the following circumstances: (Include a brief explanation for your choice in each part). a. A 2.00 ml sample of vinegar was used instead of the 1.00 mL b. A 0.100 M NaOH is used as the titrant rather than 0.050 M c. The vinegar sample was dissolved in 35 mL of H20 and not 25 mL
titration lab report

additional images of lab report

Answer 1

1. Given concentration of NaOH = 0.05 M (N_{​​​​​​2})

Weight of unknown acid taken = 0.878 g (w_{​​​​​​1}​​​​)

Amount of solvent used to dissolve the given amount of acid = 100 mL (v_{​​​​​​1}​​​​​)

Amount of acid solution used for titration = 10 mL (V^{​​​​​​}_{​​​​​​1})

Amount of NaOH required to obtain end point for 10 mL of acid solution = 18.3 mL (V_{​​​​​​2})

Let the molar mass of the unknown acid be m.

Now, for the reaction to achieve the end point, the milliequivalent of acid solution should be equal to that of the base, or

N1V1 = N2V2

N 4 = NzV2 00 x 1000 x y z , V- m б• $18 х 11р. у 10 2 О.06 18•3 37:g 2 O •OS X 18, 2 та 81'g o- 0 5 x 18.3 - 25 - 056а 4 мля"

So the molar mass of unknown acid is found to be 95.956 g/mol.

2.

2. The same equation of mulluqurnalence is to be followed here, ie, D NINI = NzV2 to here normality where No concentration of minegar W = moland (M) No volume of ueneger takin y none of Nash & volume of NaOH taken Initially, Vi= 1 mL N = 0.05M

NOW, V = 2 mL so, NIN = N₂ by Lux V₂ " N & Ny are comilant) 3 As the volume of to am , the volume increase wineger increased of tibrant would from 1 ml also b. let so N, V, = constant Ng U₂ = cenitant. Laos (N=m) volume of titrant would inverely depend on its concentration en Ar concentration increased from oo5M to 0.1m, the volume of thrant would decrease. C . dissolved Vinegar naniple distilled with in 35 m2 tho instead of 25 ml. we know, la N,V, = N₂ V => N L V / concentration of - w x 1000 Luzun titrant (N2) e volume mi of vineger w are - I a forned) w ginen weight of vinegen m molar meight of vineger o volume of water it u diroled in Intitially, U= 25 m 2 volume of water increases volo, u= 35 ml 3 ve

to As volienne of mater in inneruely related volume of titrant, the volume of tikant mould decrease