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IMG_0247.jpegIMG_0247.jpeg

Giải cho tôi câu 46

Python

python
from PIL import Image, ImageEnhance
img=Image.open('/mnt/data/IMG_0247.jpeg')
img.size


(1242, 1569)

Python

python
import matplotlib.pyplot as plt, numpy as np
arr=np.array(img)
# crop around question 46 area; approximate y range
crop=img.crop((70, 1090, 1170, 1490))
crop.size


(1100, 400)

Python

python
import matplotlib.pyplot as plt
plt.figure(figsize=(10,4))
plt.imshow(crop)
plt.axis('off')
plt.show()

Python

python
import math
Ka=2e-4; C=0.1
x=(-Ka+math.sqrt(Ka**2+4*Ka*C))/2
x, -math.log10(x)


(0.004373253849269008, 2.3591953129279397)

Python

python
C=0.1/3
x=(-Ka+math.sqrt(Ka**2+4*Ka*C))/2
pH=-math.log10(x)
C,x,pH


(0.03333333333333333, 0.002483924663504466, 2.604861580283081)

Python

python
import mpmath as mp
Ka=2e-4
Kw=1e-14
V=1.02
F=0.1/V
Na=0.005/V
F,Na


(0.09803921568627451, 0.004901960784313725)

Python

python
def f(logh):
    h=10**logh
    # compute a from Ka equation: a = Ka*(F-a)/h -> a = Ka*(F-a)/h
    # solve for a: a = Ka*(F-a)/h => a = Ka*F/h - Ka*a/h => a*(1+Ka/h)=Ka*F/h => a = Ka*F/(h+Ka)
    a=Ka*F/(h+Ka)
    oh=Kw/h
    return h+Na - (a+oh)
# find root in log10 space
for guess in [-3,-2.5,-2,-1.5]:
    try:
        root= mp.findroot(lambda x: f(x), guess)
        print(guess, root)
    except: pass

Python

python
h=10**(-2.60862435340406)
a=Ka*F/(h+Ka)
h,a, -math.log10(h), a/(F-a), math.log10(a/(F-a))+(-math.log10(Ka))


(0.002462496640040745,
 0.007364457420293622,
 2.60862435340406,
 0.08121838493013772,
 2.60862435340406)

Python

python
HA0=0.095/1.02
A0=0.005/1.02
# solve x using Ka equation
import sympy as sp
x=sp.Symbol('x', positive=True)
sol=sp.solve(sp.Eq(Ka, (x*(A0+x))/(HA0-x)), x)
sol


[0.00246249663896245]

Pensamiento

Finalizing calculations and clarifying assumptions

I’ve found that x=0.0024625, which matches the calculated [H+], leading to a pH of 2.6086. For part d, the final answer is 2.61. In part c, the missing volume likely implies a 1 L initial volume, so when adding 2 L of water, the total volume becomes 3 L. I’ll specify that assumption clearly.

Final answer and computation for part d

Let’s summarize the steps for each part: a) pKa = 3.70
b) For C = 0.1M, solving the equation gives pH = 2.36.
c) After dilution, pH = 2.60.
d) After mixing, using stoichiometry and Ka calculation, x = 0.00246, leading to a pH of 2.61. For part d, we’ll show the quadratic solution with the equation: x = [-(A0 + Ka) + sqrt((A0 + Ka)^2 + 4Ka * HA0)]/2.

Thought for 2m 32s

Respuesta

Câu 46: Ka(HCOOH)=2⋅10−4K_a(\mathrm{HCOOH}) = 2\cdot 10^{-4}

a) Tính pKapK_a

pKa=−log⁡Ka=−log⁡(2⋅10−4)=3.70pK_a=-\log K_a=-\log(2\cdot 10^{-4})=3.70

b) Tính pH dung dịch HCOOH 0,1 M0{,}1\,\mathrm{M}

Đặt [H+]=x[H^+]=x. Khi đó:

Ka=x2C−x(C=0.1)⇒x2+Kax−KaC=0K_a=\frac{x^2}{C-x}\quad (C=0.1) \Rightarrow x^2+K_ax-K_aC=0 x=−Ka+Ka2+4KaC2=−2⋅10−4+(2⋅10−4)2+4(2⋅10−4)(0.1)2≈4.37⋅10−3x=\frac{-K_a+\sqrt{K_a^2+4K_aC}}{2} =\frac{-2\cdot10^{-4}+\sqrt{(2\cdot10^{-4})^2+4(2\cdot10^{-4})(0.1)}}{2} \approx 4.37\cdot 10^{-3} pH=−log⁡x≈−log⁡(4.37⋅10−3)≈2.36\mathrm{pH}=-\log x \approx -\log(4.37\cdot 10^{-3}) \approx 2.36

c) Thêm 2 lít nước vào dung dịch axit 0,1M0{,}1\mathrm{M} → tính pH

(Thường bài ngầm hiểu ban đầu có 1 lít dung dịch 0.1M0.1\mathrm{M}.)

  • Ban đầu: V1=1 LV_1=1\,\mathrm{L}, C1=0.1 MC_1=0.1\,\mathrm{M}
  • Thêm 2 L2\,\mathrm{L} nước ⇒ V2=3 LV_2=3\,\mathrm{L}
C2=C1V1V2=0.1⋅13=0.0333 MC_2=C_1\frac{V_1}{V_2}=0.1\cdot\frac{1}{3}=0.0333\,\mathrm{M}

Làm như câu (b):

x=−Ka+Ka2+4KaC22≈2.48⋅10−3⇒pH≈2.60x=\frac{-K_a+\sqrt{K_a^2+4K_aC_2}}{2}\approx 2.48\cdot 10^{-3} \Rightarrow \mathrm{pH}\approx 2.60

d) Thêm 20 mL20\,\mathrm{mL} NaOH 0,25M0{,}25\mathrm{M} → tính pH

(giả sử thêm vào 1 lít HCOOH 0.1M0.1\mathrm{M})

Bước 1: phản ứng trung hoà

nHCOOH=0.1⋅1=0.10 moln_{\mathrm{HCOOH}}=0.1\cdot 1=0.10\ \text{mol} nOH−=0.25⋅0.020=0.005 moln_{\mathrm{OH^-}}=0.25\cdot 0.020=0.005\ \text{mol} OH−+HCOOH→HCOO−+H2O\mathrm{OH^-}+ \mathrm{HCOOH}\rightarrow \mathrm{HCOO^-}+ \mathrm{H_2O}

Sau phản ứng:

  • nHCOOH=0.10−0.005=0.095 moln_{\mathrm{HCOOH}}=0.10-0.005=0.095\ \text{mol}
  • nHCOO−=0.005 moln_{\mathrm{HCOO^-}}=0.005\ \text{mol}
  • V=1.020 LV=1.020\,\mathrm{L}

Nồng độ “ban đầu sau trộn”:

[HCOOH]0=0.0951.02=0.09314 M,[HCOO−]0=0.0051.02=0.004902 M[\mathrm{HCOOH}]_0=\frac{0.095}{1.02}=0.09314\,\mathrm{M},\quad [\mathrm{HCOO^-}]_0=\frac{0.005}{1.02}=0.004902\,\mathrm{M}

Bước 2: cân bằng axit yếu Gọi x=[H+]x=[H^+] do HCOOH phân li thêm:

Ka=x([HCOO−]0+x)([HCOOH]0−x)K_a=\frac{x([\mathrm{HCOO^-}]_0+x)}{([\mathrm{HCOOH}]_0-x)}

Thế số giải ra:

x≈2.46⋅10−3⇒pH=−log⁡x≈2.61x\approx 2.46\cdot 10^{-3}\Rightarrow \mathrm{pH}=-\log x\approx 2.61

Kết quả:

  • a) pKa≈3.70pK_a \approx 3.70
  • b) pH≈2.36\mathrm{pH}\approx 2.36
  • c) pH≈2.60\mathrm{pH}\approx 2.60 (nếu ban đầu 1 L rồi thêm 2 L nước)
  • d) pH≈2.61\mathrm{pH}\approx 2.61 (nếu thêm NaOH vào 1 L dung dịch 0.1 M)

Nếu đề của bạn không phải 1 lít ban đầu ở (c) và (d), nói mình thể tích ban đầu là bao nhiêu, mình sẽ thay lại số ra đúng ngay.

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