Answer:
Explanation:
Equation of the reaction:
NaOH + HCl --> NaCl + H2O
Volume of HCl = 5 ml
Molar concentration = 1 M
Number of moles = molar concentration * volume
= 1 * 0.005
= 0.005 mol of HCl
By stoichiometry, 1 mole of HCl completely neutralizes 1 mole of NaOH
Therefore, number of moles of NaOH = 0.005 mol
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
NaOH --> Na+ + OH-
Mass = molar mass * number of moles
= 40 * 0.005
= 0.2 g of Na+
<u>Answer:</u> The pH of the solution is 9.71
<u>Explanation:</u>
1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions.
We are given:
pOH of the solution = 7.2
To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOh of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=5.09\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D5.09%5Ctimes%2010%5E%7B-5%7DM)
Putting values in above equation, we get:
To calculate pH of the solution, we use the equation:
Hence, the pH of the solution is 9.71
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
Molecular mass of H2O is: 16+2*1=18
so moles of 40g of H2O is: 40/18=2.22
