Asolution of household vinegar is to be analyzed. a pipet is used to measure out 10 ml of the vinegar which is placed in a 250ml volumetric flask. distilled water is added until the total volume of solution is 250 ml portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide. the neutralization reaction is as follows: hc2h3o2 (aq) + oh- (aq) --> c2h3o2- (aq) + h2o (l)it is found that 16.7 ml of 0.0500 m naoh is needed to titrate 25.0 ml of the diluted vinegar. calculate the molarity of the diluted vinegar.

0.0344 M

Explanation:

Because one mol of vinegar (acetic acid) reacts with one mol of NaOH, we can use the formula

Where C₁ and V₁ refer to the concentration and volume of vinegar, and C₂ and V₂ to those of NaOH. We're given V₁, C₂ and V₂; so we solve for C₁:

the formula would be mgso4

mgso[subscript]4

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