Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is 
.
Hence, this is the required solution.
Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.
So first, we take our measured value, .299 cm, minus our known value, .225 cm.
.299 cm - .225 cm=.004 cm
Next, we divide that by our known value
Finally, multiply your answer by 100
.0177777778 x 100= 1.77777778 %
Round to three significant figures, and you're done.
=1.78 % error
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Hope this helps!
https://youtu.be/MT5tKijKZwQ
