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bezimeni [28]
3 years ago
13

How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration

Chemistry
1 answer:
Elena L [17]3 years ago
4 0

Answer:

Amount of HCL = 0.00318 L  of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH  = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH  = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed =  0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795  / 2.5

Amount of HCL = 0.00318 L  of 3.18 ml

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Answer:

False it is H2O

Explanation:

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What is the mass of oxygen in 300 grams of carbonic acid (H2CO3)
DENIUS [597]

mass of carbonic acid = 300g

molar mass of H2CO3 = 2H + C + 3 O

= 2 x 1.008+ 12.01 + 3 x  16

= 62.03g/mol


moles of H2CO3 = mass/Molar mass

= 300/62.03

= 4.8364 moles


1 mole H2CO3 has 3 moles Oxygen


4.8364 moles H2CO3 contains  

=   3 x 4.8364  moles Oxygen  =   14.509 moles Oxygen


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mass of oxygen in 300g of carbonic acid(H2CO3) = 232.15g

7 0
3 years ago
Calculate the concentrations of h2so3, hso−3, so2−3, h3o+ and oh− in 0.025 m h2so3.
sammy [17]
We will use this two reaction equation:

H2SO3 + H2O ↔ H3O+  +  HSO3-    Ka1 = 1.3 x 10^-2

HSO3-  + H2O ↔ H3O+   + SO3 2-    Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

              H2SO3 + H2O ↔ H3O+ +  HSO3- 

initial     0.025                        0            0

change   -X                             +X          +X

Equ       (0.025-X)                     X             X 

 
Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X
                   
 ∴[H3O+] = 0.0127 M


and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

                 = 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

            = 8.66 x 10^-13 M

- by using the ICE table for the second equation:

              HSO3- + H2O ↔ H3O+         + SO3 2-

initial    0.0127                      0.0127            0

change    -X                            +X                +X

Equ      (0.0127-X)                (0.0127+X)        X


when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X) 

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X 

∴[SO32-] = 6.3 x 10^-8
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