chem question?

Question

1. Benzoic acid, C6H5COOH, dissociates in water. C6H5COOH (s) ---> C6H5COO-(aq) + H+(aq) ka= 6.46x10^-5. A 25.0ml sample of an aqueous soln of pure benz. acid it titrated usin standardized 0.150M NaOH
a.) after addition of 15.0ml of the 0.150M NaOH, the pH of the soln= 4.37 calculate the following:
* [H+] in soln
* [OH-] in soln
* the # of moles of NaOH added
* The # of moles of C6H5COO- (aq) in soln.
* The # of moles of C6H5COOH in soln.
b.) State whether the soln at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning

In a different titration, 0.7529g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in water and titrated with 0.150M NaOH. The equivalence point is reached at 24.78mL of the base soln is added
c.) calculate each of the following:
* The mass, in grams of benzoic acid in the solid sample.
* The mass percentage of benzoic acid in the solid sample.

Answer 1

Question 1: Amount of NaOH (moles) = VxM where V is volume in L and M is molarity. So our moles of NaOH= 15x10-3 L x 0.15 M = 2.25x10-3 Molar.

H+ at pH=4.37 =10-5 x 10^ -0.63 = 3.8x10-5 appx.
Since [H+][OH-]= 10x10-15, [OH]- = 2.6 x 10-10 appx.
The titration rxn is BeCOOH + NaOH -> H2O + BeCOO- . We can envision the reaction as starting with 2.25x10-3 M benzoate and an unknown amount of acid. However, since the mix is still acid, we assume that the amount of acid is still about what it initially was minus that used to form the benzoate. The benzoate
Then [2.25x10-3 ][3.8x10-5] /(a-2.25x10-3)=
6.46x10-5
Now 2.25x10-3/(a-2.25x10-3) = 1.7 appx.
Then the BeCOOH STILL IN SOLUTION is about 1.3x10-3 moles appx
The equivalence point is basic, since a weak acid is being titrated with a strong base.

2. The moles of NaOH= 0.02478 x 0.15 = 0.0037
The weight of 0.0037 moles of BeCOOH is 122x0.0037= 0.5 g appx.
The sample is about 66% BeCOOH by weight

Answer 2

a million.0 Potassium is very electropositive ingredient with low ionization ability. This factors makes the outer valance electrons ejection fairly basic with electric voltage. As you already know electric impulses are via bypass of electrons 2.0 4 moles of Al reacts with 3 moles O2 giving Aluminum oxide. this skill that 8 moles of Al want six moles of O2 3.0 team III era iv, probable the ingredient is Ga and the Oxide is Ga2O3 4.0 1s2 2s2 2p6 3s13Px1