Chlorine= 35.5 +35.3= 71 a.m.u.
When pure HA is added to the buffer, the buffer component ratio and the pH decrease.
<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>
When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.
The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.
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Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
The concentration of the sodium chloride would be 0.082 M
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.
Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles
Equivalent mole of NaCl = 0.325 moles.
Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M
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