Achemistry student weighs out 0.306 g of citric acid (h3c6h5o7), a triprotic acid, into a 250 ml volumetric flask and dilutes to the mark with distilled water. he plans to titrate the acid with 0.1000 m naoh solution. calculate the volume of naoh solution the student will need to add to reach the final equivalence point. be sure your answer has the correct number of significant digits.

The volume of NaOH solution is "47.8 mL".

Given values are:

Concentration of NaOH ,

Mass,

The equation,

→

Now,

→ The moles of will be:

=

=

=

→ Moles of NaOH will be:

=

=

=

hence,

→ The volume of NaOH will be:

=

=

=

or,

=

Thus the above response is right.

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Explanation:

For simplicity, let's write the formula of citric acid as H₃A.

1. Balanced chemical equation.

2. Moles of H₃A

3. Moles of NaOH.

4. Volume of NaOH

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol