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algol [13]
4 years ago
14

Calculate the pH of a buffer solution prepared by mixing 200 mL of 0.10 M NaF and 100 mL of 0.050 M HF.

Chemistry
1 answer:
dlinn [17]4 years ago
8 0

Answer:

pH=3.74

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one could compute the pH considering that the pKa of hydrofluoric acid, HF, is 3.14:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base and acid are computed by considering the mixing process with a total volume of 300 mL (0.3 L):

n_{HF}=0.1L*0.05mol/L=0.005molHF\\\\n_{NaF}=0.2L*0.1mol/L=0.02molNaF

[HF]=\frac{0.005molHF}{0.3L}=0.017M

[NaF]=\frac{0.02molHF}{0.3L}=0.067M

Therefore, the pH turns out:

pH=3.14+log(\frac{0.067M}{0.017M} )\\\\pH=3.74

Regards.

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How many atoms are there in 0.32g of copper
aivan3 [116]

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        <u><em>calculation</em></u>

<u><em>    </em></u>Step  1: find the number of moles  of Copper

       moles =  mass/molar  mass

      =  0.32  g /63.5  g/mol=0.005  moles

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7 0
3 years ago
it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine
Allisa [31]

Answer:

The wavelength of light require to brake an single I-I bond is  7.92 × 10⁻⁷ m

Explanation:

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amount of energy to break one iodine -iodine bond = (151 KJ/mol )/ 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ    

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c = speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ  

λ   = hc / E

λ   =  (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ   = 19.878 × 10⁻²⁶ j .m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

6 0
3 years ago
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