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what the ph of 0.109 M hydrogen chromate ion (HCrO4^-)

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Chemistry

what the ph of 0.109 M hydrogen chromate ion (HCrO4^-)

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3.8 Ratings (428 Votes)

Answer - We are given, [HCrO4-] = 0.109 M ,

We know Ka2 for the H2CrO4 = 3.2*10-7

we need to put ICE table for calculating the [H3O+]

    HCrO4- + H2O ------> H3O+ + CrO42-

I 0.109                          0           0

C   -x                            +x          +x

E 0.109-x                       +x          +x

Ka = [H3O+] [CrO42-] / [HCrO4-]

3.2*10-7 = x*x /(0.005-x)

3.2*10-7 *(0.109-x) = x2

The Ka value is too small, so we can neglect the x in the 0.109-x.

3.2*10-7*0.109 =x2

x = 0.000187 M

so, x = [H3O+] = 0.000187 M

so, pH = -log [H3O+]

           = -log 0.000187 M

           = 3.73

So answer for this question is 3.73
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