0.250g of phthalic acid is dissolved in 40.00cm^3 of 0.100 mol dm^-3 NaOH. It requires 11.80 cubic centimeters of 0.100 mol\dm^-3 HCL fo neutralisation( to phenolphthalein) The mole ratio in which phthalic acid reacts with sodium hydroxide ihn 1:2

(a) Calculate the moles of excess NaOH reacted with HCL
(b)Calculate the moles of phthalic acid used
(c) Calculate the molecular mass of phthalic acid

p.s-urgently need help to solve this before my lab today :'( (22.09.11)

Phthalic acid = H2T

H2T + 2NaOH ==> Na2T + 2H2O
(a1)moles NaOH initially = M x L = 0.1 x 0.004000 L = ??
(a2)moles HCl reacted = 0.0118 x 0.100M = ??
(a3)mole excess NaOH = moles NaOH initially-moles HCl (a2-a1)=a3
(b)moles H2T used = 1/2 moles NaOH used
(c)molar mass = 0.250g/moles H2T. I get something like 166. The actual molar mass is about 177.

hey thanx heaps for your help much appreciated :)

CHEERS

(a) To calculate the moles of excess NaOH reacted with HCl, we need to determine the moles of HCl reacted first.

Using the equation NaOH + HCl → NaCl + H2O, we can see that the mole ratio between NaOH and HCl is 1:1.

Moles of HCl = concentration (mol/dm^3) × volume (dm^3)
Moles of HCl = 0.100 mol/dm^3 × 11.80 × 10^-3 dm^3
Moles of HCl = 0.00118 mol

Since the mole ratio of NaOH to HCl is 1:1, the moles of excess NaOH reacted is also 0.00118 mol.

(b) The mole ratio of phthalic acid to NaOH is 1:2.

Moles of NaOH used = concentration (mol/dm^3) × volume (dm^3)
Moles of NaOH used = 0.100 mol/dm^3 × 40.00 × 10^-3 dm^3
Moles of NaOH used = 0.004 mol

Since the mole ratio of phthalic acid to NaOH is 1:2, the moles of phthalic acid used is half of the moles of NaOH used.
Moles of phthalic acid used = 0.004 mol ÷ 2 = 0.002 mol

(c) To calculate the molecular mass of phthalic acid, we need to use its molar mass.

The molar mass of phthalic acid (C8H6O4) is:
C = 12.01 g/mol (8 atoms)
H = 1.01 g/mol (6 atoms)
O = 16.00 g/mol (4 atoms)

Molar mass of phthalic acid = (12.01 g/mol × 8) + (1.01 g/mol × 6) + (16.00 g/mol × 4)
Molar mass of phthalic acid = 96.08 g/mol + 6.06 g/mol + 64.00 g/mol
Molar mass of phthalic acid = 166.14 g/mol

Therefore, the molecular mass of phthalic acid is 166.14 g/mol.

To solve this problem, we need to follow a few steps. Let's go through them one by one.

(a) Calculate the moles of excess NaOH reacted with HCl:
First, we need to find the moles of NaOH used in the neutralization reaction with HCl. We know that the concentration of NaOH is 0.100 mol/dm^3 and the volume used is 11.80 cm^3. To find the moles, we can use the formula:

moles = concentration × volume

moles of NaOH = 0.100 mol/dm^3 × 11.80 cm^3

However, we need to convert the volume from cm^3 to dm^3. Since 1 dm^3 = 1000 cm^3, we have:

moles of NaOH = 0.100 mol/dm^3 × (11.80 cm^3 / 1000 cm^3)

Calculate the result to find the number of moles of NaOH reacted.

(b) Calculate the moles of phthalic acid used:
Now we can use the mole ratio provided to calculate the moles of phthalic acid used. The ratio is 1:2, meaning for every mole of phthalic acid, two moles of NaOH react.

moles of phthalic acid = (moles of NaOH reacted) / 2

Use the moles of NaOH calculated in step (a) and divide it by 2 to find the moles of phthalic acid used.

(c) Calculate the molecular mass of phthalic acid:
To calculate the molecular mass of phthalic acid, we need to know the mass and the number of moles. We know that the mass is 0.250 g, and we have already calculated the moles of phthalic acid used in step (b).

Molecular mass (g/mol) = mass of phthalic acid (g) / moles

Calculate the result to find the molecular mass of phthalic acid.

Now, follow these steps to solve your problem and find the answers to parts (a), (b), and (c). Good luck with your lab!