1.5M 500ml ÀÇ pH 11ÀÎ ¿ë¾×À» pH 8.8·Î ¸¸µé±â À§Çؼ
¿©±â¼ HCLÀÇ ¼øµµ´Â 30% ÀÔ´Ï´Ù.
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120,000¸®ÅÍÀÇ pH5ÀÎ ¹°À» pH6.5·Î ¸¸µé±â À§Çؼ
pH 5 = -log[H+] ..... pH 5ÀÏ ¶§ [H+] ³óµµ = 10^(-5) g-mole/L
pH 6.5 = -log[H+] ..... pH6.5ÀÏ ¶§ [H+] ³óµµ = 10^(-6.5) g-mole/L
µû¶ó¼ pH5¿Í pH6.5°£ÀÇ [H+]ÀÇ ³óµµÂ÷ ¡â[H+] = (10^-5) - (10^-6.5) g-mole/L
NaOH·Î ÁßÈÇÏ¿© ¾ø¾îÁø ¾çÀº = 9.68377E-06
(ÁßÈ¿¡ ÇÊ¿äÇÑ NaOH·®,g-mole) = 120,000L * 9.68377E-06 g-mole/L = 1.162 g-mole
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ÀÌ ½Ä´ë·Î Àû¿ëÀ» ÇÏÀÚ´Ï. ¸¶À̳ʽº °ªÀÌ ³ª¿Í¼¿ä. ¤Ì
pH 11 = -log[H+] ..... pH 11ÀÏ ¶§ [H+] ³óµµ = 10^(-11) g-mole/L
pH 8.8 = -log[H+] ..... pH8.8ÀÏ ¶§ [H+] ³óµµ = 10^(-8.8) g-mole/L
µû¶ó¼ pH11¿Í pH8.8°£ÀÇ [H+]ÀÇ ³óµµÂ÷ ¡â[H+] = (10^-11) - (10^-8.8) g-mole/L
NaOH·Î ÁßÈÇÏ¿© ¾ø¾îÁø ¾çÀº = -1.57488E-09
(ÁßÈ¿¡ ÇÊ¿äÇÑ NaOH·®,g-mole) = 0.5L * -1.57488E-09 g-mole/L = -7.8744*10ÀÇ -10½Â g-mole
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