pH of the solution after 24. 00 ml of the hcl has been added is 12.87
millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL
.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40......-2.40.........+2.40.... +2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)
pOH = -log (OH^-). Then
pOH = -log (0.075)
pOH =1.1249
As we know,
pH + pOH = pKw = 14.00
pH=14-pOH
pH=14-1.1249
pH=12.87
<h3>
What is pH?</h3>
pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.
The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.
Learn more about pH:
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Answer:
A
Explanation:
No temperature change was observed, hence the change is neither exothermic nor endothermic. Hence the answer is A.
Answer:
The pH is 7.54
Explanation:
The Henderson - Hasselbalch equation states that for a buffer solution which consists of a weak acid and its conjugate base, the buffer pH is given by:
pH ![=pk_{a} +log(\frac{[conjugate base]}{[weakacid]})](https://tex.z-dn.net/?f=%3Dpk_%7Ba%7D%20%2Blog%28%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bweakacid%5D%7D%29)
pkₐ is for the acid
In this case, the buffer hypochlorous acid HClO is a weak acid, and its conjugate base is the hypochlorite anion ClO⁻ is delivered to the solution via sodium hypochlorite NaClO
.
NaCIO = 0.200 M
HCIO = 0.200 M
pkₐ = -log₁₀ kₐ = -log₁₀ (2.9 × 10⁻⁸) = 7.54
∴pH = 
= 7.54
Answer:
32.8%
Explanation:
All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).
First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:
- 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄
There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.
We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:
- 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb
Finally we calculate the percentage composition of Pb:
- 0.083 g Pb / 0.254 g salt * 100% = 32.8%
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