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Nikitich [7]
3 years ago
11

Potassium hydrogen phthalate, KHP, is used as a primary standard for determining the concentration of a solution of NaOH by titr

ation. If the KHP has not been dried before weighing, the calculated molarity of the NaOH would be
A) higher than the actual value, since water is included in the apparent mass of KHP

B) higher than the actual value, since the presence of water requires a larger volume of titrant

C) lower than the actual value, since NaOH absorbs water

D) unaffected, since KHP is a strong acid

E) unaffected, since water is routinely added before the titration
Chemistry
1 answer:
prisoha [69]3 years ago
6 0

Answer:A

Explanation:

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2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)
defon

Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

                       207.2 g of Pb ----------------- 1 mol

                             x g             ----------------- 0.0013 moles

                        x = (0.0013 x 207.2) / 1

                        x = 0.269 g of Pb                                                                

8 0
4 years ago
Read 2 more answers
Calculate the pH for each case in the titration of 50.0 mL of 0.230 M HClO ( aq ) 0.230 M HClO(aq) with 0.230 M KOH ( aq ) . 0.2
Vikentia [17]

Answer:

pH before addition of KOH = 4.03

pH after addition of 25 ml KOH = 7.40

pH after addition of 30 ml KOH = 7.57

pH after addition of 40 ml KOH = 8.00

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pH after addition 0f 60 ml KOH = 12.3

Explanation:

pH of each case in the titration given below

(6) After addition of 60 ml KOH

Since addition of 10 ml extra KOH is added after netralisation point.

Concentration of solution after addition 60 ml KOH is calculated by

M₁V₁ = M₂V₂

or, 0.23 x 10 = (50 + 60)ml x M₂

or M₂ = 0.03 Molar

so, concentration of KOH = 0.03 molar

                               [OH⁻] = 0.03 molar

                                 pOH = 0.657

                                  pH = 14 - 0.657 = 13.34

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Answer:

HCl + Ca(OH)2 = CaCl2 + H2O - Chemical Equation Balancer.

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Mixture

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