The pH of 1 MPO 43- aq solution is-Given -pK b PO 43-1-62-A- 13-19B- 1-62C- 8-1D- 4-86

[Given pK_b (PO^3 _4) = 1.62], [PO^3 _4]=1 MPO_4^3 + H_2O ⇌ HPO^2 _4 + OH^ [OH^ ] = √(K _bC) we know pOH = 12[pK_b log C] pOH= 12[1.62 log 1] = 0.81 Also, ...

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Standard XII

Chemistry

Properties of Water

The pH of 1 M...

Question

The pH of $1 M P O_{4}^{3 -}$ (aq) solution is:
[Given $p K_{b} (P O_{4}^{3 -}) = 1.62]$

13.19

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1.62

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8.1

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4.86

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Solution

The correct option is A 13.19
[Given $p K_{b} (P O_{4}^{3 -}) = 1.62], [P O_{4}^{3 -}] = 1 M$ $P O_{4}^{3 -} + H_{2} O ⇌ H P O_{4}^{2 -} + O H^{-}$
$[O H^{-}] = \sqrt{K_{b} C}$
we know
$p O H = \frac{1}{2} [p K_{b} - l o g C]$
$p O H = \frac{1}{2} [1.62 - l o g 1] = 0.81$
Also,
$p H = 14 - p O H$
$p H = 14 - 0.81 = 13.19$

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Similar questions

Q. Calculate pH of

1 M P O_{4}^{3 -}

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p^{k_{b}} (P O_{4}^{3 -}) = 1.62

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Q. The pH of

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[Given

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Properties of Water

Standard XII Chemistry

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The pH of 1 MPO3−4 (aq) solution is: [Given pKb(PO3−4)=1.62]

The correct option is A 13.19[Given pKb(PO3−4)=1.62], [PO3−4]=1 MPO3−4+H2O⇌HPO2−4+OH− [OH−]=√KbC we know pOH=12[pKb−log C] pOH=12[1.62−log 1]=0.81 Also, pH=14−pOH pH=14−0.81=13.19

The pH of $1 M P O_{4}^{3 -}$ (aq) solution is:
[Given $p K_{b} (P O_{4}^{3 -}) = 1.62]$