35.0 ml of acid with an unknown concentration is titrated with 24.6 ml of 0.432 m base. what is the concentration of the acid? assume the acid contributes 1 mole of (h+) ions/mole of acid and the base contributes 1 mole of (oh-) ions/mole of base.

3.29 m
0.615 m
0.304 m

Given:

35.0 mL of acid with an unknown concentration

24.6 mL of 0.432 M base

Required:

Concentration of the acid

Solution:

M1V1 = M2V2

M1 (35.0 mL of acid) = (0.432 M base) (24.6 mL of base)

V1 = (0.432 M base) (24.6 mL of base) / (35.0 mL of acid)

M1 = 0.304 M of acid

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