A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH3CO2) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places.

Answer : The pH at equivalence is, 9.21

Explanation : Given,

Concentration of = 0.5870 M

Volume of = 90.0 mL = 0.09 L (1 L = 1000 mL)

First we have to calculate the moles of

As we known that at equivalent point, the moles of and NaOH are equal.

So, Moles of NaOH = Moles of = 0.0528 mol

Now we have to calculate the volume of NaOH.

Total volume of solution = 0.09 L + 0.0253 L = 0.1153 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

Moles of = 0.0528 mol

At equivalent point,

Given:

Now put all the given values in the above expression, we get:

Therefore, the pH at equivalence is, 9.21

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2