How many grams of NaHCO3 would have to be added to 2.00 L of 0.100 M H2CO3 to yield a solution with a pH = 6.00? (Ka = 4.2x10-7
pH=pKa+log[HCO2−][H2CO3]pH=pK_a+log\dfrac{[HCO_2^-]}{[H_2CO_3]}pH=pKa+log[H2CO3][HCO2−]
6=6.38+log[HCO3−]0.16=6.38+log\dfrac{[HCO_3^-]}{0.1}6=6.38+log0.1[HCO3−]
[HCO3−]=0.042 M[HCO_3^-]=0.042\;M[HCO3−]=0.042M
m=0.042 M∗2L∗84.01(gmole)=7.06 g of NaHCO3m=0.042\;M*2L*84.01(\frac{g}{mole})=7.06\;g\;of\;NaHCO_3m=0.042M∗2L∗84.01(moleg)=7.06gofNaHCO3
7.06 grams of NaHCO3 should be added.
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