titration calculation quesion (1 Viewer)

[karen88]

i have problems with this question.... could anyone help me?

A household cleaning agent contains a weak baseof general formula NaX. 1.00g of this compound was dissolved in 100.00mL of water. A 20.0 mL sample of the solution was titrated with 0.1000 mol/L hydrochloric acid and required 24.4 mL of the acid for neutralisation.

what is the molar mass of this base?

 

[LaCe]

i think this works

moles of HCl= 0.1*0.0244
= 0.00244 mol
moles of NaX= 0.00244 mol
0.00244 = mass/molar mass
molar mass = mass/moles
where mass is the original mass of NaX

 

[Dreamerish*~]

not fully sure.
i thought that strong bases generally have metals in the compound? Na should form a strong base.

NaX + HCl -> NaCl + HX ... [equation]

1.00g/100.0mL

20.0mL was titrated. therefore 1.00g x 0.2 = 0.2g of NaX was titrated.

24.4mL of HCl was required.
C = 0.1m L^-1
= 0.00244m/24.4mL

the ratio of NaX to HCl is 1:1
therefore 0.00244m of NaX was used

0.00244m = 0.2g
1m = 81.967g

therefore molar mass of NaX = 81.967g

 

[Pace_T]

.1 * 0.0244
= moles of HCL used

= moles of NaX used

mp = .2 g
(10g per litre *.02)

.'. .2g / 0.0244 = 81.97 = 82g per mol
(mp / moles = molar mass)

 

[Pace_T]

1 g put in 100mL
1g per 0.1L

.'. therefore there is 10g of solid per litre
20ml is chosen

there is 50 20mL's in a 1 litre solution
so therefore 1 50th of the 10g in in the 20ml solution
1/50 * 10g = .2g

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ok nevermind, i see you worked it out