Answer:
c. chloroacetate ion
Explanation:
The chloroacetic acid, ClCH₂CO₂H, is a weak acid with Ka = 1.36x10⁻³. When this weak acid is in solution with its conjugate base, ClCH₂CO₂⁻ (From sodium chloroacetate) a buffer is produced. The addition of a strong acid as the HCl produce the following reaction
HCl + ClCH₂CO₂⁻ → ClCH₂CO₂H + Cl⁻.
Where the acid reacts with the chloroacetate ion to produce more chloroacetic acid
That means, the HCl reacts with the chloroacetate ion present in the buffer solution
Right answer is:
<h3>c. chloroacetate ion</h3>
91.4 grams
91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
C = mol/volume
2.45M=mol/0.5L
2.45M⋅0.5L = mol
mol = 1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225mol * 
mol=1.225
=1.225 mol . 
=1.225 . 74.6
=91.4g
therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
What is 1 molar solution?
In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.
58.44 g make up a 1M solution of NaCl.
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Using the Bronsted-Lowry (B/L) Theory tell which is the acid and which the base in .... (c) [H+<span>] = </span>8.1x10-10M [OH-] = 1x10-14/8.1x10-10<span> = 0.12x10</span>-4<span> = 1.2x10</span>-5M ... (c) [OH-] = 4.5x10-7M [H+] = 1x10-14/4.5x10-7<span> = 0.22x10</span>-7<span> = 2.2</span>x10<span>-8</span><span>M</span>
