Help with this titration calculation!

Have I done it right?



Succinic acid has the formula (CH2)n(COOH)2 and reacts with dilute sodium hydroxide as follows: (CH2)n(COOH)2 + 2NaOH ? (CH2)n(COONa)2 + 2H2O
2.0 g of succinic acid were dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 18.4 cm3 was required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Deduce the molecular formula of the acid and hence the value of n.


My method

25*0.1/1000=0.0025 which is the moles of NaOH
0.0025/2=0.00125 which is moles of succinic acid
18.4cm^3/1000=0.0184dm^3
0.00125/0.0184= 0.068moldm^3 which is concentration of succinic acid

2g in 250cm^3 ------> Therefore there is 2g in 0.25dm^3-----> Therefore there is 8gdm^3

8/0.068=117.65 which is Mr of succinic acid
Mr of (COOH)2 = 90
117.65-90=27.65
27.65/2=1.975 which round to 2

Therefore n is 2. This is the correct answer just want to check my method. I ahve a test coming up so this is important.

Greatly appreciated!

Yeah it's fine...

There are a couple of slightly different variations, but they all amount to the same thing.

Yeah it's fine...

There are a couple of slightly different variations, but they all amount to the same thing.