The pH of a solution obtained by mixing 10mL of 0.45M HCl and 40 mL of 0.1 M NaOH is ???
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Heyy mate ❤✌✌❤
Here's your Answer...
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=>HCl + NaOH ---> NaCl + H2O
=>moles HCl = (0.04L)(0.1 mol/L) = 0.004 moles
=>moles NaOH = (0.01 L)(0.45 mol/L ) = 0.0045 moles
=>Excess NaOH = 0.0045 moles - 0.004 moles = 0.0005 moles
Total volume = 0.04 L + 0.01 L = 0.05 L
=>molarity of OH = 0.0005 moles/0.05 L = 0.01 M
=>pOH = -log [OH] = -log 0.01
=>pOH = 2
=>pH = 14 - pOH
=>pH = 12
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Here's your Answer...
⤵️⤵️⤵️⤵️⤵️⤵️⤵️
=>HCl + NaOH ---> NaCl + H2O
=>moles HCl = (0.04L)(0.1 mol/L) = 0.004 moles
=>moles NaOH = (0.01 L)(0.45 mol/L ) = 0.0045 moles
=>Excess NaOH = 0.0045 moles - 0.004 moles = 0.0005 moles
Total volume = 0.04 L + 0.01 L = 0.05 L
=>molarity of OH = 0.0005 moles/0.05 L = 0.01 M
=>pOH = -log [OH] = -log 0.01
=>pOH = 2
=>pH = 14 - pOH
=>pH = 12
✔✔✔
Anonymous:
i don't think this answer's right because it was asked in exam that it's answer would be between 0 - 9
i have done step by step. please check
still the answer has to come between 0 and 9
let me check sir
Aj u did just the opposite ...
just letting you know that answer is 2
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