Question

2. If we add 25.0 mL of 0.220 M Ca(OH)2 to 50.0 mL of 0.105 M HF, will the mixture be acidic, basic or neutral? Given: 2 HF (aq) + Ca(OH)2 (aq) + CaF2 (aq) + 2 H20 m) What are the pH, POH, [H30*), and [OH-] values for the solution formed? 0.220m Ca(OH)2 25mL 50ml 0.105m HF
can someone please explain step by step

Answer 1

+ Calonlan Ca fsi)t 2H20) 2HF(9) for his Solution ferned to be completety neutsal Fer erery one mete t Ca(a4)2(2) two metes f HF() equire giren t 0.930M 0.220 mefe/Lifre CaloH)2 Lit X o.2 20 mote Lit 40 25 ale f caloH), TH 1000 SMK10mele 17 and 0-105 Mele/ Lit 6.L05M= HF -3 - 52Sx(0 mete mate ot HE= 50 /000 it ~3 mote ef Cal oH Ca) fer erery s- SXIO sequires txjo-3 mote of HE peutral. but there onty s.25 KiOmote be for so Lo 6o1 noiu be Basic. CaEs (91)t 2 H, O(e) + CaloH)ela2) 2HF() 5-2510met 5.5y103mt 0 atto -3 5-5X10- 2 e at tote s.25%2 binle, at ern mete 1tF 0 5 25 K103-2x S0, .25 K10 -3 -3 2-625X/0 mete

at e 50 caton mele -e caleH)= Ca2e# 55x10 2-625I03 mete Cal oH),= 2met - (S.-2.625)/o-3 2-875x/0- mete = 2x.875x/omete 7x(0-3 Lit Concontintien Co 25 mtt 50me =75 mi so1 2x 2- 75 7 5 mele Lit 2x3.8 3 x102mot/ut COH- 7.66X 10-2 met/ - Loy Lota - Lo7 7-60810-27 POH = 1LIS poH= Sin ce 14 Pit +poit 50 PH 14-1115 2.885 Pit= Sinte - L70047 - 70 0M PH 12-885 -12-885 -/3 1303X10 CH77