pH = - lg c(H+)ã
pH = 12ï¼å³c(H+) = 1Ã10^(-12)mol/Lï¼èc(OH-) = Kw / c(H+) = 0.01mol/Lã
pH = 11ï¼å³c(H+) = 1Ã10^(-11)mol/Lï¼èc(OH-) = Kw / c(H+) = 0.001mol/Lã
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¥10mLï¼å 为KOH溶äºæµåº¦å¾å°ï¼å¯åº¦ä¸æ°´æ¥è¿ï¼
ï¼2ï¼è®¾éNaOH溶液xmLï¼[10Ã10^(-3)LÃ0.01mol/L + xÃ10^(-3)LÃ0.0001mol/L] ÷ [10Ã10^(-3)L + xÃ10^(-3)L] = 0.001mol/Lï¼è§£å¾ï¼x = 100ã
å¦æå å
¥pH = 10ï¼ä¹å³c(OH-) = 0.0001mol/LçNaOH溶液ï¼éå å
¥100mLã
ï¼3ï¼è®¾éH2SO4溶液ymLï¼[10Ã10^(-3)mLÃ0.01mol/L - yÃ10^(-3)mLÃ0.001mol/LÃ2] ÷ [10Ã10^(-3)mL + yÃ10^(-3)mL] = 0.001mol/Lï¼è§£å¾ï¼y = 30ã
å¦æå å
¥0.001mol/LçH2SO4溶液ï¼éå å
¥30mLã
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