At certain temperature T if the conductivity of pure water is 5-5 - 10-7Scm-1- then calculate pH of water at given temperature-Given - -H-350Scm2sq-1- -OH-200Scm2eq-1- -OH-200 Scm2sq-1

The correct option is D 5Conductivity of pure water = 5.5 × 10^ 7 S cm^ 1 λ ^∞ _H^+ = 350 S cm^2 sq^ 1 #160; #160; #160; #160; λ ^∞ _O ...

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Standard XII

Chemistry

pH the Power of H

At certain te...

Question

At certain temperature $(T)$ if the conductivity of pure water is $5.5 \times 10^{- 7} S c m^{- 1}$ , then calculate $p H$ of water at given temperature.
Given : $λ_{H +}^{\infty} = 350 S c m^{2} s q^{- 1}; λ_{O H^{-}}^{\infty} \infty 200 S c m^{2} e q^{- 1}; λ_{O H^{-}}^{\infty} = 200 S c m^{2} s q^{- 1}$

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Solution

The correct option is D 5
Conductivity of pure water = $5.5 \times 10^{- 7} S c m^{- 1}$
$λ_{H^{+}}^{\infty} = 350 S c m^{2} s q^{- 1}$ $λ_{O H^{-}}^{\infty} = 200 S c m^{2} s q^{- 1}$
At infinity $λ = 550 S c m^{2} s q^{- 1}$
$α_{O H^{-}} = \frac{λ}{λ_{\circ}} = \frac{5.5 \times 10^{- 7}}{550} = 10^{- 9}$
$p O H = - l o g (10^{- 9}) = 9$
$p H = 14 - 9 = 5$

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At certain temperature (T) if the conductivity of pure water is 5.5×10−7Scm−1, then calculate pH of water at given temperature.Given : λ∞H+=350Scm2sq−1;λ∞OH−∞200Scm2eq−1;λ∞OH−=200Scm2sq−1

The correct option is D 5Conductivity of pure water = 5.5×10−7Scm−1λ∞H+=350Scm2sq−1 λ∞OH−=200Scm2sq−1At infinity λ=550Scm2sq−1αOH−=λλ∘=5.5×10−7550=10−9pOH=−log(10−9)=9pH=14−9=5