In buffer solution there is an equilibrium between the acid HA and its conjugate base A⁻: HA(aq) ⇌ H⁺(aq) + A⁻(aq).
When acid (H⁺ ions) is added to the buffer solution, the equilibrium is shifted to the left, because conjugate base (A⁻) reacts with hydrogen cations from added acid, according to Le Chatelier's principle: H⁺(aq) + A⁻(aq) ⇄ HA(aq). So, the conjugate base (A⁻) consumes some hydrogen cations and pH is not decreasing (less H⁺ ions, higher pH of solution).
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable
Answer:
I know that the 100-mL graduated cylinders are always read to 1 decimal place.
I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.
Answer:
Zn + 2HCl → ZnCl2 + H2
Explanation:
Zn + HCl → ZnCl2 +
The complete equation is given below:
Zn+ HCl → ZnCl2 + H2
Now we can balance the equation by doing the following:
There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:
Zn + 2HCl → ZnCl2 + H2
An ionic compound is a substance that contains atoms that are bonded together through an ionic bond, where electrons are transferred from one atom to another. For the given experiment above, the material that is missing from the list is WATER or solvent. It is necessary because conduction through water will help you identify if a sample compound is ionic through its electrolytes. <span>Ionic compounds that are soluble are typically electrolytes</span>
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
