<span>Let x = amt of distilled water
:
A simple equation
.25(16) = .10(x+16)
4 = .10x + 1.6
4 - 1.6 = .1x
2.4 = .1x
x = 2.4/0.1
x = 24 oz of distilled water
:
:
Prove this by seeing the amt of antiseptic is the same (only the % changes)
.25(16) = .10(24+16)
4 = .1(40)
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Answer:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
<em></em>
∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>
