0.2 mole of NaOH is added to one liter of a buffer solution that0.8 M NaHCO 3 an

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The ionic equation for this would be: OH- +H2CO3 -> H2O +HCO3- For this, you should use a neutralization table to help organizeyour thoughts:                       OH-         H2CO3              H2O            HCO3- Initial:          .2moles   .7moles            n/a              .8 moles Change:       -.2 moles -.2moles           n/a              +.2 moles Final:           0 moles    .5moles            n/a              1 moles From here, you can use the Henderson-Hasselbach equation, which ispH = pKa + log([A-]/[HA]). The Ka should be given in your textbook,which is 4.3e-7. The pKa is -log(Ka), which gives you 6.3665. pH = 6.3665 + log([1]/[.5]) pH = 6.66753

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