HELPPPPPPPP PLZZZZ MVP ME .!!!

A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.Part A: What is the initial pH of this solution?Part B: What

Lelechka [254]

Answer:

Initial pH of this pH = 7.453

pH after addition of 150.0 mg of HBr = 7.35

pH after addition of 85.0 mg of NaOH 0.154 = 7.56

Explanation:

Since Ka value isn't given

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log $\frac{[Conjugate base]}{[Acid]}$

pH = 7.52 + log $\frac{0.15}{0.175}$

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr

$\frac{Mass}{Molar mass} \\= \frac{150 X10^{-3}g }{80.91} \\= 0.00185 } mole$

Moles of NaOCl in 100 ml buffer solution = $\frac{0.15X100}{1000}$ = 0.015

Moles of HClO in 100 ml buffer solution = $\frac{0.175X100}{1000}$ = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

= 0.0132

moles of HClO = 0.0175 + 0.00185

= 0.0194

Using Henderson equation pH = Pka + log $\frac{Conjugate base}{Acid}$

= 7.52 + log $\frac{0.0132}{0.0194}$

= 7.35

Part C

pH after addition of 85 mg of HBr

moles of NaOH

$\frac{Mass}{Molar mass} \\= \frac{85 X10^{-3}g }{40} \\= 0.00213 } mole$

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

= 0.0171

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

= 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log $\frac{Conjugate base}{Acid}$

= 7.52 + log $\frac{0.171}{0.154}$

= 7.56

3 0

3 years ago

How do you determine the type of ion charge any element will form based on its number of valence electrons?

erastova [34]

When electrons are gained the element is negatively charged and when electrons are loss the element is positively charged

8 0

4 years ago

At which temperature would a reaction with H -92 kJ/mol, S -0.199 kJ/(mol-K) be spontaneous? A.600k B.500k C.400k D.700k

stiv31 [10]

Given information : H = -92 KJ/mol and S = -0.199 KJ/(mol.K)

At equilibrium G = 0

We have to find the Temperature at which reaction would be spontaneous.

For spontaneous reaction : $\triangle G = negative (-)$

For non-spontaneous reaction : $\triangle G = positive (+)$

We can find the temperature using the formula for Gibbs free energy which is:

$\triangle G = \bigtriangleup H - T\bigtriangleup S$

Where, G = Gibbs free energy ,

H = Enthalpy

S = Entropy

T = Temperature

By plugging the value of G , H and S in the above formula we can find 'T'

$\triangle G = \bigtriangleup H - T\bigtriangleup S$

Since reaction should be spontaneous that means $\triangle G$ should be negative , so the above formula can be written as :

$\triangle G < \bigtriangleup H - T\bigtriangleup S$

On rearranging the above formula we get :

$0 < \bigtriangleup H - T\bigtriangleup S$

$T < \frac{\bigtriangleup H}{\bigtriangleup S}$

$T < \frac{-92\frac{KJ}{mol}}{-0.199\frac{KJ}{mol.K}}$

$T < (\frac{-92}{-0.199})\times (\frac{KJ}{mol})\times (\frac{mol.K}{KJ})$

$T < 462.3 K$

For the reaction to be spontaneous , T should be less than 462.3 K, so out of given option , C is correct which is 400 K.

8 0

3 years ago

Read 2 more answers

Consider the reaction.

Alex_Xolod [135]

Note the signs of equilibrium:-

Reaction don't procede forward or backward
Concentration of products and reactants remains same .

So

if

Concentration of A is 2M then concentration of B should be same .

So equilibrium constant K is 1

$\\ \rm\rightarrowtail K=\dfrac{[Products]^a}{[Reactants]^b}$

So

[B]=[A]^2
[B]=2^2
[B]=4M

8 0

2 years ago

In a titration of 35.00 mL of 0.737 M H2SO4, __________ mL of a 0.827 M KOH solution is required for neutralization.

elena55 [62]

Answer:

In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.

Explanation:

The balanced reaction is

H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.

The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:

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

in units of $\frac{moles}{liter}$

then the number of moles can be calculated as:

number of moles= molarity* volume

You have acid H₂SO₄

35.00 mL= 0.035 L (being 1,000 mL= 1 L)
Molarity= 0.737 M

Then:

number of moles= 0.737 M* 0.035 L

number of moles= 0.0258

So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?

$moles of KOH=\frac{0.0258moles of H_{2} SO_{4}*2 moles of KOH }{1mole of H_{2} SO_{4}}$